JEE SECTION - A [MATHS - MCQ]

अतः यदि $n$ द सम है, तब उत्तर $(b)$ है तथा यदि $(n)$ विषम है, तब उत्तर $(a)$ होगा 

==> $\frac{A}{2} = \frac{\pi }{2} - \left( {\frac{{B + C}}{2}} \right)$

$\therefore$  $\cot \frac{A}{2} = \tan \left( {\frac{B}{2} + \frac{C}{2}} \right)$

==> $\frac{1}{{\tan \frac{A}{2}}} = \frac{{\tan \frac{B}{2} + \tan \frac{C}{2}}}{{1 - \tan \frac{B}{2}\tan \frac{C}{2}}}$

==> $1 - \tan \frac{B}{2}\tan \frac{C}{2} $

$= \tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{A}{2}.\tan \frac{C}{2}$

$\tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{A}{2}\tan \frac{C}{2} = 1$

अर्थात्,  $\sum {\tan \frac{A}{2}\tan \frac{B}{2} = 1} $.

दोनों तरफ वर्ग करने पर, $x + \frac{1}{x} + 2 = 4\,{\cos ^2}\theta $

$\Rightarrow$ $x + \frac{1}{x} = 4{\cos ^2}\theta  - 2$

$\Rightarrow$ $x + \frac{1}{x} = $$2(2{\cos ^2}\theta  - 1)$ $ = 2\cos 2\theta $…..$(ii)$

पुन: दोनों तरफ वर्ग करने पर,

${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}2\theta $

$\Rightarrow$ ${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}2\theta  - 2$$ = 2(2{\cos ^2}2\theta  - 1)$

$\Rightarrow$ ${x^2} + \frac{1}{{{x^2}}} = 2\cos 4\theta $…..$(iii)$

अब दोनों तरफ घन करने पर, ${\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^3} = {(2\cos 4\theta )^3}$

$\Rightarrow$ ${x^6} + \frac{1}{{{x^6}}} + 3{x^2} \times \frac{1}{{{x^2}}}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 8{\cos ^3}4\theta $

$\Rightarrow$ ${x^6} + \frac{1}{{{x^6}}} + 3\,(2\cos 4\theta ) = 8{\cos ^3}4\theta $

$ \Rightarrow {x^6} + \frac{1}{{{x^6}}} = 8{\cos ^3}4\theta  - 6\cos 4\theta $

$=  2\,(4{\cos ^3}4\theta  - 3\cos 4\theta )$

$= 2\cos 3(4\theta ) = 2\cos 12\theta $.

$\therefore \,\,{\sin ^2}\beta  = \sin \alpha \cos \alpha $

अब $\cos 2\beta  = 1 - 2{\sin ^2}\beta  = 1 - 2\sin \alpha \cos \alpha $

$ = {(\cos \alpha  - \sin \alpha )^2} = 2\,{\left( {\frac{1}{{\sqrt 2 }}\cos \alpha  - \frac{1}{{\sqrt 2 }}\sin \alpha } \right)^2}$

$ = 2{\sin ^2}\left( {\frac{\pi }{4} - \alpha } \right)$,

जो $(a)$ में दिया गया है

एवं $\cos 2\beta  = 2{\cos ^2}\left\{ {\frac{\pi }{2} - \left( {\frac{\pi }{4} - \alpha } \right)} \right\} $

$= 2{\cos ^2}\left( {\frac{\pi }{4} + \alpha } \right)$,

जो विकल्प $(b)$ में दिया गया है।

${x^2} + {y^2} = 1 + 1 + 2\,\cos \,(2A - A)$

$\therefore \,\,\,\frac{{{x^2} + {y^2} - 2}}{2} = \cos \,A$…..$(i)$

साथ ही, $\cos A + 2\,{\cos ^2}A - 1 = y$

या $(\cos A + 1)\,(2\,\cos A - 1) = y$

समीकरण $(i)$ से $\cos A$ का मान रखने पर उत्तर प्राप्त हो जाता है।

==> $\tan (\alpha + \beta ) = \frac{{\frac{1}{{1 + \frac{1}{{{2^x}}}}} + \frac{1}{{1 + {2^{x + 1}}}}}}{{1 - \frac{1}{{1 + 1/{2^x}}}\frac{1}{{1 + {2^{x + 1}}}}}}$

==> $\tan (\alpha + \beta ) = \frac{{{2^x} + {{2.2}^{x + x}} + {2^x} + 1}}{{1 + {2^x} + {{2.2}^x} + {{2.2}^{x + x}} - {2^x}}}$

==> $\tan (\alpha + \beta ) = 1$

$ \Rightarrow \alpha + \beta = \frac{\pi }{4}$.

$ = 2{\left( {\cos \theta  + \frac{1}{2}} \right)^2} - \frac{3}{2}$

अब $2{\left( {\cos \theta  + \frac{1}{2}} \right)^2} \ge 0$ सभी $\theta $ के लिए

$\therefore \,\,2{\left( {\cos \theta  + \frac{1}{2}} \right)^2} - \frac{3}{2} \ge \frac{{ - 3}}{2}$ सभी $\theta $ के लिए

$\cos 2\theta  + 2\cos \theta  \ge \frac{{ - 3}}{2}$ सभी $\theta $ के लिए

साथ ही, इस व्यंजक का अधिकतम मान $3$ है।

$\Rightarrow$ $A = {\cos ^2}\theta  + {\sin ^2}\theta .{\sin ^2}\theta $\

$\Rightarrow$ $A \le {\cos ^2}\theta  + {\sin ^2}\theta $,

$\Rightarrow$ $A \le 1$

पुन: $A = {\cos ^2}\theta  + {\sin ^4}\theta  = (1 - {\sin ^2}\theta ) + {\sin ^4}\theta $

$A = {\left( {{{\sin }^2}\theta  - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4}$

अत: $3/4 \le A \le 1$.

$\Rightarrow \sin A = - \frac{4}{5},\cos A = - \frac{3}{5}$

$\therefore $ $5\sin 2A + 3\sin A + 4\cos A$ 

$= 10\sin A\cos A + 3\sin A + 4\cos A$ 

$= 10\,\left( {\frac{{12}}{{25}}} \right) - \frac{{12}}{5} - \frac{{12}}{5} = 0$.

$ = 2\,[3\sin \theta - 4{\sin ^3}\theta ]\,[4{\cos ^3}\theta - 3\cos \theta ]$

$=24 \sin \theta \cos \theta (\sin ^2 \theta + \cos ^2 \theta) -18\sin \theta \cos \theta  -32 \sin ^2 \theta \cos^2 \theta$

$ = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3\sin 2\theta $

तुलना करने पर, $x = \sin 2\theta .$

==> $x = k\cos \theta $,

$y = k\cos \left( {\theta - \frac{{2\pi }}{3}} \right)$,

$z = k\cos \left( {\theta + \frac{{2\pi }}{3}} \right)$ 

==> $x + y + z = k\left[ {\cos \theta + \cos \left( {\theta - \frac{{2\pi }}{3}} \right) + \cos \left( {\theta + \frac{{2\pi }}{3}} \right)} \right]$

$ = k[(0) = 0$ 

$ \Rightarrow $ $x + y + z = 0$.

$\frac{2}{{\cos 2\alpha }} = \frac{{\sin \beta }}{{\cos \beta }} + \frac{{\cos \beta }}{{\sin \beta }}$

$= \frac{{{{\sin }^2}\beta + {{\cos }^2}\beta }}{{\cos \beta \sin \beta }}$

$ = \frac{1}{{\cos \beta .\sin \beta }}$ 

==> $\cos 2\alpha = \sin 2\beta $ 

==> $\cos 2\alpha $= $\cos \,\left( {\frac{\pi }{2} - 2\beta } \right)$

==> $2\alpha = \frac{\pi }{2} - 2\beta $ 

==> $2\alpha + 2\beta = \frac{\pi }{2}$ 

==> $\alpha + \beta = \frac{\pi }{4}$.

==> $\frac{1}{{\sin \theta }} = \frac{{p + q}}{{p - q}}$

योगान्तरानुपात नियम से, 

$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{p + q + p - q}}{{p + q - p + q}}$

==> ${\left\{ {\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} - \sin \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$ 

==> ${\left\{ {\frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$

==> ${\tan ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{p}{q}$ 

==> ${\cot ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{q}{p}$

नोट : $\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \sqrt {\frac{q}{p}}$ सिर्फ, यदि  $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) > 0$.

==> $a\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) + b\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = c$

$ \Rightarrow $ $a - a{\tan ^2}\theta + 2b\tan \theta = c + c{\tan ^2}\theta $

$ \Rightarrow $$ - (a + c){\tan ^2}\theta + 2b\,\tan \theta + (a - c) = 0$

$\therefore \tan \alpha + \tan \beta = - \frac{{2b}}{{ - (c + a)}} = \frac{{2b}}{{c + a}}$ .

$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ \,(2\sin {40^o}\sin 80^\circ )$

$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ (\cos 40^\circ - \cos 120^\circ )$

$ = \frac{1}{2}.\frac{{\sqrt 3 }}{2}\sin 20^\circ \left( {1 - 2{{\sin }^2}20^\circ + \frac{1}{2}} \right)$

$ = \frac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\frac{3}{2} - 2{{\sin }^2}20^\circ } \right)$

$ = \frac{{\sqrt 3 }}{8}(3\sin 20^\circ - 4{\sin ^3}20^\circ )$

$ = \frac{{\sqrt 3 }}{8}\sin 60^\circ = \frac{{\sqrt 3 }}{8}.\frac{{\sqrt 3 }}{2} = \frac{3}{{16}}$.

सूत्र ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ का प्रयोग करके हल करने पर अभीष्ट मान अर्थात् 

$K = \frac{3}{4}$ प्राप्त होता है।

$ \Rightarrow \tan \theta = \frac{{\sin \left( {\alpha - \frac{\pi }{4}} \right)}}{{\cos \left( {\alpha - \frac{\pi }{4}} \right)}} $

$\Rightarrow \tan \theta = \tan \left( {\alpha - \frac{\pi }{4}} \right)$

$ \Rightarrow \theta = \alpha - \frac{\pi }{4}$

$\Rightarrow \alpha = \theta + \frac{\pi }{4}$

अतः  $\sin \alpha + \cos \alpha = \sin \left( {\theta + \frac{\pi }{4}} \right) + \cos \left( {\theta + \frac{\pi }{4}} \right)$

$ = \sqrt 2 \cos \theta $

एवं  $\sin \alpha - \cos \alpha = \sin \left( {\theta + \frac{\pi }{4}} \right) - \cos \left( {\theta + \frac{\pi }{4}} \right)$

$ = \frac{1}{{\sqrt 2 }}\sin \theta + \frac{1}{{\sqrt 2 }}\cos \theta - \frac{1}{{\sqrt 2 }}\cos \theta + \frac{1}{{\sqrt 2 }}\sin \theta $

$ = \frac{2}{{\sqrt 2 }}\sin \theta = \sqrt 2 \sin \theta $.

$ \Rightarrow \frac{{n - 1}}{{n + 1}} = \frac{{\sin A - \sin B}}{{\sin A + \sin B}} $

$= \frac{{2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}}}{{2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2}}}$ 

$ = \tan \frac{{A - B}}{2}\cot \frac{{A + B}}{2}$ 

$ \Rightarrow \frac{{n - 1}}{{n + 1}}\tan \left( {\frac{{A + B}}{2}} \right) = \tan \frac{{A - B}}{2}$ .

$ = 2\cos (A + B)\cos (A - B) + \cos 2C$

$ = 2\cos \left( {\frac{{3\pi }}{2} - C} \right)\cos (A - B) + \cos 2C$

$ =  - 2\sin C\cos (A - B) + 1 - 2{\sin ^2}C$

$ = 1 - 2\sin C\{ \cos (A - B) + \sin C\} $

$ = 1 - 2\sin C\left\{ {\cos (A - B) + \sin \left( {\frac{{3\pi }}{2} - (A + B)} \right)} \right\}$

$ = 1 - 2\sin C\{ \cos (A - B) - \cos (A + B)\} $

$ = 1 - 4\sin A\sin B\sin C$.

ट्रिक : $A = B = C = \frac{\pi }{2}$ रखकर परीक्षण करें।

$ \Rightarrow \tan (A + B) = \tan (\pi  - C)$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan C}} = \tan (\pi  - C)$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} =  - \tan C$

अब चूँकि  $C$ एक अधिक कोण है। अत:

$ \Rightarrow \tan C < 0 \Rightarrow  - \tan C > 0$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} > 0 $

$\Rightarrow 1 - \tan A\tan B > 0$

$( \because A,B$ न्यून कोण हैं अत: $\therefore \tan A > 0,\tan B > 0)$

$ \Rightarrow \tan A\tan B < 1$.

$\therefore \,\frac{A}{2} + \frac{B}{2} = {90^o} - \frac{C}{2}$

$\therefore \cot \left( {\frac{A}{2} + \frac{B}{2}} \right) = \cot \left( {{{90}^o} - \frac{C}{2}} \right)$

या $\frac{{\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\, - 1}}{{\cot \frac{B}{2}\, + \,\cot \frac{A}{2}}} = \tan \frac{C}{2} = \frac{1}{{\cot \frac{C}{2}}}$

या $\left( {\cot \frac{A}{2}\cot \frac{B}{2} - 1} \right)\cot \frac{C}{2} = \cot \frac{B}{2} + \cot \frac{A}{2}$

$\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\,.\,\cot \frac{C}{2} = \cot \frac{C}{2} + \cot \frac{B}{2}$ $ + \cot \frac{A}{2}.$

$ = \frac{{\sin (B + C)}}{{\sin B\,\sin C}}$

$ = \frac{{\sin ({{180}^o} - A)}}{{\sin B\,\sin C}}$

$ = \frac{{\sin A}}{{\sin B\sin C}}$

इसी प्रकार, $\cot C + \cot A = \frac{{\sin B}}{{\sin C\sin A}}$

एवं $\cot A + \cot B = \frac{{\sin C}}{{\sin A\sin B}}$

अत: $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$

$ = \frac{{\sin A}}{{\sin B\sin C}}.\frac{{\sin B}}{{\sin C\sin A}}.\frac{{\sin C}}{{\sin A\sin B}}$

$ = \cos {\rm{ec}}A\cos {\rm{ec}}B\cos {\rm{ec}}C.$ 

$A + B + C = \pi  \Rightarrow B + C = \pi  - A$

 $\therefore \cos (B + C) = \cos (\pi  - A) \Rightarrow \cos (B + C) =  - \cos A$

$ \Rightarrow \cos B\cos C - \sin B\sin C =  - \cos B\cos C$

$( \because {\rm{Given}}\cos A = \cos B\cos C)$ 

 $ \Rightarrow 2\cos B\cos C = \sin B\sin C$

$ \Rightarrow \frac{{\cos B\cos C}}{{\sin B\sin C}} = \frac{1}{2}$

$\Rightarrow \cot B\cot C = \frac{1}{2}$.

$= \frac{{{S_1} - {S_3}}}{{1 - {S_2}}} = \tan \frac{\pi }{2} = \infty $ 

$\therefore {S_2} = 1$ or $xy + yz + zx = 1$, 

यहाँ  $x = \tan \frac{A}{2}$ इत्यादि. 

अब ${(x - y)^2} + {(y - z)^2} + {(z - x)^2} \ge 0$

या $2\sum {x^2} - 2\sum xy \ge 0 \Rightarrow \sum {x^2} \ge 1$.   $\{ \because \sum xy = 1\} $

$\therefore \tan A\tan B\tan C = 6 \Rightarrow \tan C = \frac{6}{2} = 3$

एवं $\tan A + \tan B = 6 - 3 = 3$

$\therefore \tan A,\tan B = 2,1$ या  $1, 2$ तथा  $\tan C = 3$.

$ = 1 - {\cos ^2}A + 1 - {\cos ^2}B + {\sin ^2}C$

$ = 2 - {\cos ^2}A - \cos (B + C)\cos (B - C)$

$ = 2 - \cos A[\cos A - \cos (B - C)]$

$ = 2 - \cos A[ - \cos (B + C) - \cos (B - C)]$ 

$ = 2 + \cos A.2\cos B\cos C$  

$\therefore$ ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\cos B\cos C = 2$.

एवं ${N^r} = 4\sin A\sin B\sin c$ 

$\therefore L.H.S. = \frac{{{N^r}}}{{{D^r}}}$

तथा $\sin A = 2\sin \frac{A}{2}\cos \frac{A}{2}$ .

$ = - 1 - 2\cos C\cos (A - B) + 2{\cos ^2}C$ 

$ = - 1 - 2\cos C[\cos (A - B) + \cos (A + B)]$

$ = - 1 - 4\cos A\cos B\cos C$.

तब  $N = 3\sin \theta - 4{\sin ^3}\theta - (4{\cos ^3}\theta - 3\cos \theta )$ 

$ = 3(\sin \theta + \cos \theta ) - 4({\sin ^3}\theta + {\cos ^3}\theta )$ 

$ = (\sin \theta + \cos \theta )\{ 3 - 4({\sin ^2}\theta - \sin \theta \cos \theta + {\cos ^2}\theta )\} $ 

$\therefore \;\frac{N}{D} + 1 $

$=  \frac{{(\sin \theta + \cos \theta )\{ 3 - 4(1 - \sin \theta \cos \theta )\} }}{{\sin \theta + \cos \theta }} + 1$

$ = 3 - 4(1 - \sin \theta \cos \theta ) + 1$

$ = 4\sin \theta \cos \theta = 2\sin 2\theta $.

$ = \frac{{\sin A - \cos A + 1}}{{\sin A - 1 + \cos A}} $

$= \frac{{\sin A + (1 - \cos A)}}{{\sin A - (1 - \cos A)}}$

$ = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2} + 2{{\sin }^2}\frac{A}{2}}}{{2\sin \frac{A}{2}\cos \frac{A}{2} - 2{{\sin }^2}\frac{A}{2}}}$

$ = \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}} $

$= \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}^2}}}{{{{\cos }^2}\frac{A}{2} - {{\sin }^2}\frac{A}{2}}}$

$ = \frac{{1 + \sin A}}{{\cos A}}$.

अब $\theta  = \phi   = \frac{\pi }{4}$ रखने पर,

$\cos 2\left( {\frac{\pi }{2}} \right) - 4\cos \left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{4}} \right) + 2{\sin ^2}\left( {\frac{{2\pi }}{4}} \right) = 0$

विकल्प $(a)$ में $\theta  = \phi = \frac{\pi }{4}$ रखने पर, $\cos 2\theta  = \cos \frac{\pi }{2} = 0$

अत: विकल्प $(a)$ सही है।

एवं $\cos 2\,\theta  + \cos 2\,\phi  = 3/2$…..$(ii)$

वर्ग करके जोड़ने पर,

$\therefore \,({\sin ^2}2\theta  + {\cos ^2}2\theta ) + ({\sin ^2}2\phi  + {\cos ^2}2\phi  )$

$ + 2\,[\sin 2\,\theta \,\sin 2\,\phi  + \cos 2\,\theta \,\cos 2\,\phi  ] = (1/4) + (9/4)$

$\Rightarrow$ $\cos 2\theta \cos 2\,\phi  + \sin 2\theta \sin 2\phi  = 1/4$

$\Rightarrow$ $\cos (2\theta  - 2\phi  ) = 1/4$

$\Rightarrow$  ${\cos ^2}(\theta  - \phi   ) = 5/8$.

==> $\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} = 2\cos \frac{\alpha }{2}\cos \frac{\beta }{2} - 2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$

$ \Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2} = \cos \frac{\alpha }{2}\cos \frac{\beta }{2}$ 

==> $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{1}{3}$.

==> $\tan A = \frac{3}{2}\tan B = \frac{3}{2}t$,   [माना $\tan B = t$] 

==> $\sin 2B = \frac{{2t}}{{1 + {t^2}}},\cos 2B = \frac{{1 - {t^2}}}{{1 + {t^2}}}$ 

$\therefore$ $\frac{{\left( {\frac{{2t}}{{1 + {t^2}}}} \right)}}{{5 - \left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right)}}$

$ = \frac{{2t}}{{4 + 6{t^2}}} = \frac{t}{{2 + 3{t^2}}} = \tan (A - B)$.

$\Rightarrow$ $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$

अत:  $\sqrt {1 + \sin A}  = \sin \frac{A}{2} + \cos \frac{A}{2}$…..$(i)$

तथा $\sqrt {1 - \sin A}  = \sin \frac{A}{2} - \cos \frac{A}{2}$…..$(ii)$

समी $(i)$ में से $(ii)$ को घटाने पर,

$2\cos \frac{A}{2} = \sqrt {1 + \sin A}  - \sqrt {1 - \sin A} $.

$\therefore$ $25{\cos ^2}\alpha  + 5\cos \alpha  - 12 = 0$

$\cos \alpha  = \frac{{ - 5 \pm \sqrt {25 + 1200} }}{{50}}$ $ = \frac{{ - 5 \pm 35}}{{50}}$

$\cos \alpha  =  - 4/5$                   $[ \because \pi /2 < \alpha  < \pi  \Rightarrow \cos \alpha  < 0]$ 

$\therefore $$\sin 2\alpha  = 2\sin \alpha \cos \alpha  =  - 24/25$.

$= \frac{{\sqrt 2 - \sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } \right\}}}{{\sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } \right\}}}$

$=\frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} \right)}}$

$= \frac{{\sqrt 2 \left\{ {\,1 - \cos \theta } \right\}}}{{\sqrt 2 \sin \theta }},$  जहाँ  $\theta = \alpha - \frac{\pi }{4}$

$= \frac{{2{{\sin }^2}(\theta /2)}}{{2\sin (\theta /2)\cos (\theta /2)}} = \tan \frac{\theta }{2}$ 

$ = \tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$.

$\sin 2\theta  = {x^2} - 1 \le 1 \Rightarrow {x^2} \le 2$

या  $ - \sqrt 2  \le x \le \sqrt 2 $          $[\because \,\,\sin 2\theta  \le 1]$

अब ${\sin ^6}\theta  + {\cos ^6}\theta $

$ = {({\sin ^2}\theta  + {\cos ^2}\theta )^3} - 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta  + {\cos ^2}\theta )$

$ = 1 - 3{\sin ^2}\theta {\cos ^2}\theta  = 1 - \frac{3}{4}{\sin ^2}2\theta $

$ = 1 - \frac{3}{4}{({x^2} - 1)^2} = \frac{1}{4}\{ 4 - 3{({x^2} - 1)^2}\} $

अत: दिया गया परिणाम सत्य होगा जब ${x^2} \le 2$ ना कि $x$ की सभी वास्तविक संख्याओं के लिए.

$\sin 2\beta = \frac{{2T}}{{1 + {T^2}}} = \frac{3}{5} \Rightarrow \cos 2\beta = \frac{4}{5}$   {$T = \tan \beta $}

$\therefore \,\,\sin 4\beta = 2\sin 2\beta \cos 2\beta $

$= 2.\frac{3}{5}.\frac{4}{5} = \frac{{24}}{{25}} = \cos 2\alpha $.

$ = 16(2{\cos ^2}A - 1 - 4{\cos ^3}A + 3\cos A)$ 

$ = 16\left( {2 \times \frac{9}{{16}} - 1 - 4 \times \frac{{27}}{{64}} + 3 \times \frac{3}{4}} \right) = 11$.

$\Rightarrow$ $\tan 2\beta  = \frac{{2\tan \beta }}{{1 - {{\tan }^2}\beta }} = \frac{3}{4}$

$\therefore \tan (\alpha  + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$

अब $0 < \beta  < \frac{\pi }{2}$ एवं $\tan 2\beta  = \frac{3}{4} > 0$ दोनों

$\Rightarrow$  $0 < 2\beta  < \frac{\pi }{2}$.

पुन: $0 < \alpha  < \frac{\pi }{2}$ एवं $0 < 2\beta  < \frac{\pi }{2}$ दोनों 

$\Rightarrow$  $0 < \alpha  + 2\beta  < \pi $

अत: $0 < \alpha  + 2\beta  < \pi $ एवं $\tan (\alpha  + 2\beta ) = 1$ दोनों     

$\Rightarrow$  $\alpha  + 2\beta  = \frac{\pi }{4} $

$\Rightarrow 2\beta  = \frac{\pi }{4} - \alpha $.

तब  $\frac{{4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2{{\cos }^2}\frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = y$

==> $\frac{{2\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2} + \sin \frac{\alpha }{2}}} \times \frac{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}}{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}} = y$

==> $\frac{{1 - \cos \alpha + \sin \alpha }}{{1 + \sin \alpha }} = y$.

इसलिए  $\cos (\theta  - \phi ) = \cos \theta \cos \phi + \sin \theta \sin \phi $

$ = \frac{3}{5}.\frac{4}{5} + \frac{4}{5}.\frac{3}{5} = \frac{{24}}{{25}}$

लेकिन $2{\cos ^2}\left( {\frac{{\theta  - \phi }}{2}} \right) $

$= 1 + \cos (\theta  - \phi ) = 1 + \frac{{24}}{{25}} = \frac{{49}}{{50}}$

$\therefore$ ${\cos ^2}\left( {\frac{{\theta  - \phi }}{2}} \right) = \frac{{49}}{{50}}$. 

अत: $\cos \left( {\frac{{\theta  - \phi }}{2}} \right) = \frac{7}{{5\sqrt 2 }}$.

$=  \frac{{\frac{{\sin {{70}^o}}}{{\cos {{70}^o}}} - \frac{{\sin {{20}^o}}}{{\cos {{20}^o}}}}}{{\frac{{\sin {{50}^o}}}{{\cos {{50}^o}}}}}$

$=  \frac{{\frac{{\sin {{70}^o}\cos {{20}^o} - \cos {{70}^o}\sin {{20}^o}}}{{\cos {{70}^o}\cos {{20}^o}}}}}{{\frac{{\sin {{50}^o}}}{{\cos {{50}^o}}}}}$

$=  \frac{2}{2} \times \frac{{\sin ({{70}^o} - {{20}^o})\cos {{50}^o}}}{{\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\sin {{50}^o}\cos {{50}^o}}}{{2\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\cos {{50}^o}}}{{\cos {{90}^o} + \cos {{50}^o}}}$

$= \frac{{2\cos {{50}^o}}}{{0 + \cos {{50}^o}}}$

$= 2.$

$\frac{{(\cos \,6x + \cos \,4x) + 5\,(\cos \,4x + \cos \,2x) + 10\,(\cos \,2x + 1)}}{{\cos \,5x + 5\,\cos \,3x + 10\,\cos x}}$

सरल करने पर अभीष्ट परिणाम अर्थात्, $2\,\cos x$ है।

$ = \tan \,\,{9^o} - \tan \,\,{27^o} - \cot \,\,{27^o} + \cot \,\,{9^o}$

$ = (\tan \,\,{9^o} + \cot \,\,{9^o}) - (\tan \,\,{27^o} + \cot \,\,{27^o})$

$ = \frac{{\cos ({9^o} - {9^o})}}{{\sin {9^o}\cos {9^o}}} - \frac{{\cos ({{27}^o} - {{27}^o})}}{{\sin {{27}^o}.\cos {{27}^o}}} = \frac{2}{{\sin {{18}^o}}} - \frac{2}{{\sin {{54}^o}}}$

$ = 2\,\left\{ {\frac{{\sin \,\,{{54}^o} - \sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}\,\sin \,\,{{54}^o}}}} \right\} $

$= 2.\,\,\frac{{2\,.\,\cos \,\,{{36}^o}.\,\sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}.\,\sin \,\,{{54}^o}}} = 4$

$ = \frac{1}{{2\,\,\sin \,\,{{10}^o}}}\,[2\,\,\sin \,\,{10^o}\cos \,\,{10^o}\cos \,\,{20^o}\,\,\cos \,\,{40^o}]$

$ = \frac{1}{{2\,.\,2\,\,\sin \,\,{{10}^o}}}\,[2\,\,\sin \,\,{20^o}\cos \,\,{20^o}\,\,\cos \,\,{40^o}]$

$ = \frac{1}{{2\,.\,4\sin {{10}^o}}}[2\sin {40^o}\cos {40^o})$

$= \frac{1}{{8\sin {{10}^o}}}(\sin {80^o})$

$ = \frac{1}{{8\,\,\sin \,\,{{10}^o}}}\cos \,\,{10^o} = \frac{1}{8}\cot \,\,{10^o}$.

$ = \frac{{\cos {{70}^o} + 2\sin {{140}^o}}}{{\sin {{70}^o}}} $

$= \frac{{\cos {{70}^o} + 2\sin ({{180}^o} - {{40}^o})}}{{\sin {{70}^o}}}$

$ = \frac{{\sin {{20}^o} + \sin {{40}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}} $

$= \frac{{2\sin {{30}^o}\cos {{10}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}}$

$ = \frac{{\sin {{80}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}} $

$= \frac{{2\sin {{60}^o}\cos {{20}^o}}}{{\sin {{70}^o}}} = \sqrt 3 $.

$2\sin 2\theta \cos \theta  + \sin 2\theta  = \sin \alpha $

$\sin 2\theta (2\cos \theta  + 1) = \sin \alpha $…..$(i)$

अब $\cos \theta  + \cos 3\theta  + \cos 2\theta  = \cos \alpha $

$2\cos 2\,\theta \cos \,\theta  + \cos 2\theta  = \cos \alpha $

$\cos 2\theta \,(2\cos \theta  + 1) = \cos \alpha $…..$(ii)$

समीकरण $(i)$ एवं $(ii)$ से,

$\tan 2\theta  = \tan \alpha $

$\Rightarrow$  $2\theta  = \alpha $

$\Rightarrow$  $\theta  = \alpha /2$.

दिये गये समीकरण इस प्रकार लिख सकते हैं

$\cos x + \cos y =  - \cos \alpha $ एवं $\sin x + \sin y =  - \sin \alpha $

अत: $2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) =  - \cos \alpha $…..$(i)$

$2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) =  - \sin \alpha $…..$(ii)$

समी. $(i)$ को समी. $(ii)$ से भाग देने पर,

 $\frac{{2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right)}}{{2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right)}}$

$ = \frac{{\cos \alpha }}{{\sin \alpha }}$

$\Rightarrow$ $\cot \left( {\frac{{x + y}}{2}} \right) = \cot \alpha $.

$ = \tan x + \frac{{\tan x + \sqrt 3 }}{{1 - \sqrt 3 \,\tan x}} + \frac{{\tan x - \sqrt 3 }}{{1 + \sqrt 3 \,\tan x}}$

$ = \tan x + \frac{{8\tan x}}{{1 - 3{{\tan }^2}x}} $

$= \frac{{3\,(3\tan x - {{\tan }^3}x)}}{{1 - 3{{\tan }^2}x}} = 3\tan 3x$

अत: दिया गया समीकरण 

$\Rightarrow$ $3\tan 3x = 3$==> $\tan 3x = 1.$ है।

$ = {\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{\pi }{8}$

$ = 2\left( {{{\sin }^2}\frac{\pi }{8} + {{\sin }^2}\frac{{3\pi }}{8}} \right) = 2 \times 1 = 2$.

व $\cos \theta  + \cos \phi  = b$…..$(ii)$

वर्ग करने पर, ${\sin ^2}\theta  + {\sin ^2}\phi  + 2\sin \theta \sin \phi  = {a^2}$

व ${\cos ^2}\theta  + {\cos ^2}\phi  + 2\cos \theta \cos \phi  = {b^2}$

जोड़ने पर, $2+ 2$ $(\sin \theta \sin \phi  + \cos \theta \cos \phi  ) = {a^2} + {b^2}$

$\Rightarrow$ $2\cos (\theta  - \phi   ) = {a^2} + {b^2} - 2$

$\Rightarrow$   $\cos (\theta  - \phi  ) = \frac{{{a^2} + {b^2} - 2}}{2}$

$ \Rightarrow \frac{{1 - {{\tan }^2}\frac{{\theta  - \phi  }}{2}}}{{1 + {{\tan }^2}\frac{{\theta  - \phi  }}{2}}} = \frac{{{a^2} + {b^2} - 2}}{2}$

$\Rightarrow$ $({a^2} + {b^2}) + ({a^2} + {b^2}){\tan ^2}\frac{{\theta  - \phi  }}{2} - 2 - 2{\tan ^2}\frac{{\theta  - \phi  }}{2}$

$ = 2 - 2{\tan ^2}\frac{{\theta  - \phi  }}{2}$

$\Rightarrow$ $\frac{{4 - {a^2} - {b^2}}}{{{a^2} + {b^2}}} = {\tan ^2}\frac{{\theta  - \phi  }}{2}$

$\Rightarrow$  $\tan \frac{{(\theta  - \phi  )}}{2} = \sqrt {\frac{{4 - {a^2} - {b^2}}}{{{a^2} + {b^2}}}} $ 

ट्रिक : $\theta  = \frac{\pi }{2},\phi  = {0^o}$ रखने पर, $a = 1 = b$

$\therefore$ $\tan \frac{{\theta  - \phi   }}{2} = 1$, जो विकल्प $(a)$ व $(b)$ देते हैं।

अत: पुन: $\theta  = \frac{\pi }{4} = \phi  $ रखने पर, $\tan \frac{{\theta - \phi  }}{2} = 0$ जो $(b)$ देता है।

$ = \frac{{\sin {{20}^o}}}{{\cos {{20}^o}}} - \frac{{\sin {{70}^o}}}{{\cos {{70}^o}}} + 2\tan {50^o}$

$ = \frac{{\sin {{20}^o}\cos {{70}^o} - \cos {{20}^o}\sin {{70}^o}}}{{\cos {{20}^o}\cos {{70}^o}}}$$ + 2\tan {50^o}$

$ = \frac{{\sin ({{20}^o} - {{70}^o})}}{{\frac{1}{2}[\cos ({{70}^o} + {{20}^o}) + \cos ({{70}^o} - {{20}^o})]}}$$ + 2\tan {50^o}$

$ = \frac{{2\sin ( - {{50}^o})}}{{\cos {{90}^o} + \cos {{50}^o}}} + 2\tan {50^o}$

$ = \frac{{ - 2\sin {{50}^o}}}{{0 + \cos {{50}^o}}} + 2\tan {50^o}$

$ = - 2\tan {50^o} + 2\tan {50^o} = 0$.

$ = {\cos ^2}\alpha + {\left\{ {\cos \,(\alpha + {{120}^o}) + \cos \,(\alpha - {{120}^o})} \right\}^2}$$ - 2\,\cos \,(\alpha + {120^o})\,\cos \,(\alpha - {120^o})$

$ = {\cos ^2}\alpha + {\left\{ {\,2\,\cos \,\,\alpha \,\cos \,{{120}^o}} \right\}^2} - 2\,\left\{ {{{\cos }^2}\alpha - {{\sin }^2}\,{{120}^o}} \right\}$

$ = {\cos ^2}\alpha + {\cos ^2}\alpha - 2\,{\cos ^2}\alpha + 2\,{\sin ^2}\,{120^o}$

$ = 2{\sin ^2}{120^o} $

$= 2 \times \frac{3}{4} $

$= \frac{3}{2}$.

$ = \frac{1}{2}\left[ {1 + \cos {{152}^o} + 1 + \cos {{32}^o} - \cos {{92}^o} - \cos {{60}^o}} \right]$

$ = \frac{1}{2}\left[ {2 - \frac{1}{2} + \cos {{152}^o} + \cos {{32}^o} - \cos {{92}^o}} \right]$

$ = \frac{1}{2}\left[ {\frac{3}{2} + 2\cos {{92}^o}\cos {{60}^o} - \cos {{92}^o}} \right]$

$ = \frac{1}{2}\left[ {\frac{3}{2} + \cos {{92}^o} - \cos {{92}^o}} \right]$

$ = \frac{3}{4}$.

$ = \frac{1}{4}\left[ {2\sin \frac{\pi }{{16}}\sin \frac{{3\pi }}{{16}}.2\sin \frac{{5\pi }}{{16}}\sin \frac{{7\pi }}{{16}}} \right]$

$ = \frac{1}{4}\left[ {\left( {\cos \frac{\pi }{8} - \cos \frac{\pi }{4}} \right)\left( {\cos \frac{\pi }{8} - \cos \frac{{3\pi }}{4}} \right)} \right]$

$ = \frac{1}{4}\left[ {\left( {\cos \frac{\pi }{8} - \frac{1}{{\sqrt 2 }}} \right)\left( {\cos \frac{\pi }{8} + \frac{1}{{\sqrt 2 }}} \right)} \right]$

$ = \frac{1}{4}\left[ {\left( {{{\cos }^2}\frac{\pi }{8} - \frac{1}{2}} \right)} \right] $

$= \frac{1}{8}\left[ {2{{\cos }^2}\frac{\pi }{8} - 1} \right]$

$ = \frac{1}{8}\left[ {\cos \frac{\pi }{4}} \right] $

$= \frac{1}{8} \times \frac{1}{{\sqrt 2 }}$

$= \frac{{\sqrt 2 }}{{16}}$.

$ = 1 - {\sin ^2}\left( {\frac{\pi }{{12}}} \right) + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + {\cos ^2}\left( {\frac{{5\pi }}{{12}}} \right)$

$ = 1 + \frac{1}{2} + \left( {{{\cos }^2}\frac{{5\pi }}{{12}} - {{\sin }^2}\frac{\pi }{{12}}} \right)$

$ = \frac{3}{2} + \cos \left( {\frac{{5\pi }}{{12}} + \frac{\pi }{{12}}} \right)\cos \left( {\frac{{5\pi }}{{12}} - \frac{\pi }{{12}}} \right) $

$= \frac{3}{2} + \cos \frac{\pi }{2}\cos \frac{\pi }{3}$

$ = \frac{3}{2} + 0.\frac{1}{2} = \frac{3}{2}$.

$ = \frac{1}{4}\,(2\,\,\sin \,\,{12^o}\,\sin \,\,{48^o})\,\,(2\,\,\sin \,\,{24^o}\,\,\sin \,\,{84^o})$

$ = \frac{1}{2}(\cos \,\,{36^o} - \cos \,\,{60^o})\,\,(\cos \,\,{60^o} - \cos \,\,{108^o})$

$ = \frac{1}{4}\,\left( {\cos \,\,{{36}^o} - \frac{1}{2}} \right)\,\,\left( {\frac{1}{2} + \sin \,\,{{18}^o}} \right)$

$ = \frac{1}{4}\left\{ {\frac{1}{4}(\sqrt 5 + 1) - \frac{1}{2}} \right\}\,\left\{ {\frac{1}{2} + \frac{1}{4}(\sqrt 5 - 1)} \right\} = \frac{1}{{16}}$

एवं $\cos \,\,{20^o}\,\cos \,\,{40^o}\,\,\cos \,\,60\,\,\cos \,\,{80^o}$

$ = \frac{1}{2}[\cos \,({60^o} - {20^o})\,\cos \,\,{20^o}\,\cos \,({60^o} + {20^o})]$

$ = \frac{1}{2}\,\left[ {\frac{1}{4}\cos \,\,3\,\,({{20}^o})} \right] = \frac{1}{8}\cos \,\,{60^o} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{{16}}$.

$ \Rightarrow \,\,\frac{{m + 1}}{{m - 1}} = \frac{{\cos A + \cos B}}{{\cos A - \cos B}}$

$= \frac{{2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{B - A}}{2}} \right)}}{{2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{B - A}}{2}} \right)}}$

$ = \cot \,\left( {\frac{{A + B}}{2}} \right)\,\cot \,\left( {\frac{{B - A}}{2}} \right)$

अत: $\cot \,\left( {\frac{{A + B}}{2}} \right) $

$= \frac{{m + 1}}{{m - 1}}\tan \frac{{B - A}}{2}$.

$ = 2\cos \frac{\pi }{{13}}.\cos \frac{{9\pi }}{{13}} + 2\cos \frac{{4\pi }}{{13}}\cos \frac{\pi }{{13}}$

$ = 2\cos \frac{\pi }{{13}}\left[ {\cos \frac{{9\pi }}{{13}} + \cos \frac{{4\pi }}{{13}}} \right]$ 

$ = 2\cos \frac{\pi }{{13}}\left[ {2\cos \frac{\pi }{2}.\cos \frac{{5\pi }}{{26}}} \right] = 0$,.                  $\left[ \because {\cos \frac{\pi }{2} = 0} \right]$

==> $\tan ({70^o} - {20^o}) = \frac{{\tan {{70}^o} - \tan {{20}^o}}}{{1 + \tan {{70}^o}\tan {{20}^o}}}$

==> $\tan {50^o} + \tan {70^o}\tan {20^o}\tan {50^o} = \tan {70^o} - \tan {20^o}$

==> $\tan {50^o} + \tan {50^o} = \tan {70^o} - \tan {20^o}$

                                               $[\,\because \tan {70^o} = \cot {20^o}]$

==> $2\tan {50^o} + \tan {20^o} = \tan {70^o}$

==> $2\tan {50^o} + \tan {20^o} = \tan {50^o} + \sec {50^o}$.

$ \Rightarrow \cos \alpha  =  - \frac{8}{{17}},\sin \beta  =  - \frac{{12}}{{13}}$ एवं $\cos \beta  =  - \frac{5}{{13}}$

$\pi  < \beta  < \frac{{3\pi }}{2}$ ,

$\therefore \cos \beta  =  - \frac{5}{{13}}$]

$\sin (\beta  - \alpha ) = \sin \beta \cos \alpha  - \cos \beta \sin \alpha $ = $\frac{{171}}{{221}}$.

$= 2 sin 54^\circ cos 7^\circ - 2 sin 18^\circ cos 7^\circ$

$ = \,\,2\,\,\cos \,\,{7^o}\,(\sin \,\,{54^o} - \sin \,\,{18^o})$

$ = \,\,2\,\,\cos \,\,{7^o}\,\,.\,\,2\,\,\cos \,\,{36^o}\,\,.\,\,\sin \,\,{18^o}$

$ = \,\,4.\,\cos \,\,{7^o}.\,\frac{{\sqrt 5 + 1}}{4}.\frac{{\sqrt 5 - 1}}{4} = \cos \,\,{7^{o.}}$.

$ \Rightarrow \,\,\sin \theta  = \sqrt {1 - \frac{{{8^2}}}{{{{17}^2}}}}  = \frac{{15}}{{17}}$

दिये गये व्यंजक का मान

$ = \cos \,\,{30^o}\,.\,\cos \theta  - \sin \,\,{30^o}\sin \theta  + \cos \,\,{45^o}\cos \theta $

 $ + \sin \,\,{45^o}\sin \theta  + \cos \,\,{120^o}\cos \theta  + \sin \,\,{120^o}\sin \theta $

$ = \cos \theta \,\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} - \frac{1}{2}} \right) - \sin \theta \,\left( {\frac{1}{2} - \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2}} \right)$

$ = \frac{8}{{17}}\,\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} - \frac{1}{2}} \right) + \frac{{15}}{{17}}\,\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} - \frac{1}{2}} \right)$

$ = \frac{{23}}{{17}}\,\left( {\frac{{\sqrt 3  - 1}}{2} + \frac{1}{{\sqrt 2 }}} \right)$.

$\Rightarrow \,\tan \,(A + B) = \tan \,\frac{\pi }{4}$

$ \Rightarrow \,\,\frac{{\tan A + \tan B}}{{1 - \tan A\,\tan B}} = 1$

$ \Rightarrow \,\,\tan A + \tan B + \tan A\,\tan B = 1$

$ \Rightarrow \,\,(1 + \tan A)\,(1 + \tan B) = 2$.

अब, $\cos \,(A + B) = \cos A\,\cos B - \sin A\,\sin B$

$ = \sqrt {1 - \frac{{16}}{{25}}} \,\left( { - \frac{{12}}{{13}}} \right) - \frac{4}{5}\sqrt {1 - \frac{{144}}{{169}}} $

$ =  - \frac{3}{5} \times \frac{{12}}{{13}} - \frac{4}{5}\,\left( { - \frac{5}{{13}}} \right) $

$=  - \frac{{16}}{{65}}$

(चूँकि $A$ प्रथम व $B$ तृतीय चतुर्थांश में हैं)

$= \frac{1}{{(1 + \tan A)\,(1 + \tan B)}}$ $ = \frac{1}{{\tan A + \tan B + 1 + \tan A\tan B}}$                                                                                  $[\,\because \tan (A + B) = \tan {225^o}]$

$ \Rightarrow \,\tan \,A + \tan \,B = 1 - \tan \,A\,\tan B$

$ = \frac{1}{{1 - \tan A\,\tan B + 1 + \tan A\tan B}} $

$= \frac{1}{2}$.

$\tan \,({20^o} + {40^o}) = \frac{{\tan \,{{20}^o} + \tan \,{{40}^o}}}{{1 - \tan \,{{20}^o}\,\tan \,{{40}^o}}}$

$ \Rightarrow \,\,\,\sqrt 3  = \frac{{\tan \,{{20}^o} + \tan \,{{40}^o}}}{{1 - \tan \,{{20}^o}\,\tan \,{{40}^o}}}$

$ \Rightarrow \,\,\sqrt 3  - \sqrt 3 \,\tan \,{20^o}\,\tan \,{40^o} = \tan \,{20^o} + \tan \,{40^o}$

$ \Rightarrow \,\,\tan \,{20^o} + \tan \,{40^o} + \sqrt 3 \,\tan \,{20^o}\,\tan \,{40^o} $

$= \sqrt 3 .$

$\frac{{\sin A}}{{\sin B}} = \frac{{\cos A}}{{\cos B}}\, $

$\Rightarrow \,\,\sin A\,\cos B - \cos A\,\sin B = 0$

$ \Rightarrow \,\,\sin \,(A - B) = 0$, अत: $\sin \,\left( {\frac{{A - B}}{2}} \right) = 0.$

$ \Rightarrow \,\,\frac{{2\,\,\sin \frac{{A + B}}{2}.\cos \frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2}.\cos \frac{{A - B}}{2}}} = \frac{C}{D}$

$ \Rightarrow \,\,\tan \frac{{A + B}}{2} = \frac{C}{D}$

इस प्रकार, $\sin \,(A + B) = \frac{{2\,\,\tan \frac{{A + B}}{2}}}{{1 + {{\tan }^2}\frac{{A + B}}{2}}}$

$ = \frac{{2\,\frac{C}{D}}}{{1 + \frac{{{C^2}}}{{{D^2}}}}}$

$= \frac{{2CD}}{{({C^2} + {D^2})}}$.

$= \frac{{1 - \cos \alpha + 1 + \cos \alpha }}{{\sqrt {1 - {{\cos }^2}\alpha } }}$

$ = \frac{2}{{ \pm \sin \alpha }}$

$ = \frac{2}{{ - \sin \alpha }},\,\,( $ चूँकि $\pi < \alpha < \frac{{{\rm{3}}\pi }}{{\rm{2}}} ).$

जो कि शून्य होगा यदि $\sin \left( {\frac{\pi }{4} + x - y} \right) = 0$

अर्थात्, $\frac{\pi }{4} + x - y = n\pi (n \in I) $

$\Rightarrow x = n\pi  - \frac{\pi }{4} + y$.

$ \Rightarrow \,\sin x + 3\cos x = 3\cos y - \sin y$…..$(i)$

$ \Rightarrow \,r\cos \,(x - \alpha ) = r\cos \,(y + \alpha ),$

जहाँ $r = \sqrt {10} ,\,\tan \alpha  = \frac{1}{3}$

$ \Rightarrow \,x - \alpha  =  \pm (y + \alpha )\, \Rightarrow \,x =  - y$ या $x + y = 2\alpha $

स्पष्टत: $x =  - y$ $(i)$ को सन्तुष्ट करता है

$\therefore \;\frac{{\sin \,3x}}{{\sin \,3y}} = \frac{{ - \sin \,3y}}{{\sin \,3y}} =  - 1$.

एवं $\sec (A + B) = \frac{2}{{\sqrt 3 }} \Rightarrow A + B = \frac{{11\pi }}{6}$…..(ii)

(i) व (ii) से, $B = \frac{{19\pi }}{{24}}$

==> ${\cos ^2}x + {\sin ^2}x + 2\cos x\sin x + k\sin x\cos x - 1 = 0,\,\forall \,x$

==> $(k + 2)\cos x\sin x = 0,\,\forall \,x$ ==> $k + 2 = 0$ ==> $k = - 2$.

$\therefore \,\,\,\tan \left( {\frac{{A + B}}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right)$ 

==> $\frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}.\tan \frac{B}{2}}} = \cot \frac{C}{2} $

$\Rightarrow \frac{{\frac{1}{3} + \frac{2}{3}}}{{1 - \frac{1}{3}.\frac{2}{3}}} = \frac{9}{7} = \cot \frac{C}{2}$ 

$\therefore  \,\,  \tan \frac{C}{2} = \frac{7}{9}$.

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ..... + {\sin ^2}{85^o} + {\sin ^2}{90^o}.$

हम जानते हैं कि, $\sin {90^o} = 1$ या ${\sin ^2}{90^o} = 1$.

इसी प्रकार, $\sin {45^o} = \frac{1}{{\sqrt 2 }}$ या $\,{\rm{si}}{{\rm{n}}^{\rm{2}}}{45^o} = \frac{1}{2}$

तथा $18$ पदों के कोण समान्तर श्रेणी में हैं।

हम यह भी जानते हैं${\sin ^2}{85^o} = {[\sin ({90^o} - {5^o})]^2}$$ = {\cos ^2}{5^o}$

$\therefore$ ${\sin ^2}{5^o} + {\sin ^2}{85^o} = {\sin ^2}{5^o} + {\cos ^2}{5^o} = 1.$

इसलिए ${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ... + {\sin ^2}{85^o} + {\sin ^2}{90^o}$                                                                         (पूरक नियम द्वारा)

$ = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + 1 + \frac{1}{2} = 9\frac{1}{2}$.

$\sin \theta = \frac{1}{{\sqrt {50} }},\,\,\cos \theta = \frac{7}{{\sqrt {50} }},\,\,\cos \phi = \frac{3}{{\sqrt {10} }}$

$\therefore \,\,\cos 2\phi = 2{\cos ^2}\phi - 1 = 2.\frac{9}{{10}} - 1 = \frac{8}{{10}}$

$\sin 2\phi = 2\sin \phi \cos \phi = 2 \times .\frac{1}{{\sqrt {10} }} \times \frac{3}{{\sqrt {10} }} = \frac{6}{{10}}$

$\therefore \, \cos (\theta + 2\phi ) = \cos \theta \cos 2\phi - \sin \theta \sin 2\phi $

$ = \frac{7}{{\sqrt {50} }} \times \frac{8}{{10}} - \frac{1}{{\sqrt {50} }}.\frac{6}{{10}}$

$ = \frac{{56 - 6}}{{10\sqrt {50} }} = \frac{{50}}{{10\sqrt {50} }} $

$= \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}$

$\therefore \, \theta + 2\phi = {45^o}$.

$ = 6[{({\sin ^2}\theta + {\cos ^2}\theta )^3} - 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta + {\cos ^2}\theta )]$

$ - 9[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}] + 4$

$ = 6[1 - 3{\sin ^2}\theta {\cos ^2}\theta ] - 9\,[1 - 2{\sin ^2}\theta {\cos ^2}\theta ] + 4$

$ = 6 - 9 + 4 = 1$.

एवं $\cos 22\frac{{{1^o}}}{2} = \frac{1}{2}\sqrt {2 + \sqrt 2 } $

$\therefore \left( {1 + \cos 22\frac{{{1^o}}}{2}} \right)\,\left( {1 + \cos 67\frac{{{1^o}}}{2}} \right)\,\left( {1 + \cos 112\frac{{{1^o}}}{2}} \right)$

$\left( {1 + \cos 157\frac{{{1^o}}}{2}} \right)$

$ = \left( {1 + \frac{1}{2}\sqrt {2 + \sqrt 2 } } \right)\,\left( {1 + \frac{1}{2}\sqrt {2 - \sqrt 2 } } \right)\,\left( {1 - \frac{1}{2}\sqrt {2 - \sqrt 2 } } \right)\,$

$\left( {1 - \frac{1}{2}\sqrt {2 + \sqrt 2 } } \right)$

$ = \left[ {1 - \frac{1}{4}(2 + \sqrt 2 )} \right]\,\left[ {1 - \frac{1}{4}(2 - \sqrt 2 )} \right]$

$ = \frac{{(4 - 2 - \sqrt 2 )(4 - 2 + \sqrt 2 )}}{{16}}$

$ = \frac{{(2 - \sqrt 2 )(2 + \sqrt 2 )}}{{16}} = \frac{{4 - 2}}{{16}} = \frac{1}{8}$.

एवं $\sin \theta  + \cos \theta  = b$…..$(ii)$

अब ${({b^2} - 1)^2}({a^2} + 4)$

$ = {\{ {(\sin \theta  + \cos \theta )^2} - 1\} ^2}\{ {(\tan \theta  - \cot \theta )^2} + 4\} $

$ = {[1 + \sin 2\theta  - 1]^2}[{\tan ^2}\theta  + {\cot ^2}\theta  - 2 + 4]$

$ = {\sin ^2}2\theta ({\rm{cose}}{{\rm{c}}^2}\theta  + {\sec ^2}\theta )$

$ = 4{\sin ^2}\theta {\cos ^2}\theta \left[ {\frac{1}{{{{\sin }^2}\theta }} + \frac{1}{{{{\cos }^2}\theta }}} \right] = 4$.

ट्रिक : व्यंजक ${({b^2} - 1)^2}({a^2} + 4)$ का मान $\theta $ से स्वतंत्र है इसलिए $\theta $ का कोई उपयुक्त मान रखने पर,

माना $\theta  = 45^\circ $,

अत: $a = 0,\;b = \sqrt 2 $ ताकि ${[{(\sqrt 2 )^2} - 1]^2}$ $({0^2} + 4) = 4.$

एवं $x\,\sin \,\alpha  - y\,\cos \,\alpha  = 0$…..$(ii)$

अब $(ii)$ से, $x\,\sin \,\alpha  = y\,\cos \,\alpha $

यह मान $(i)$ में रखने पर,

$ \Rightarrow \,\,y\,\cos \alpha \,{\sin ^2}\alpha  + y\,{\cos ^3}\alpha  = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \alpha \,\left\{ {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right\} = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \,\alpha  = \sin \,\alpha \,\cos \,\alpha \,$

$\Rightarrow \,\,y = \sin \,\alpha $ एवं $x = \cos \,\alpha $

अत:, ${x^2} + {y^2} = {\sin ^2}\alpha  + {\cos ^2}\alpha  = 1.$

$\Rightarrow \,\,\sin x = {\cos ^2}x$

$\therefore$ ${\cos ^8}x + 2{\cos ^6}x + {\cos ^4}x $

$= {\sin ^4}x + 2{\sin ^3}x + {\sin ^2}x$

$ = {(\sin x + {\sin ^2}x)^2} = 1$.

$\Rightarrow \,\,\cos x = {\sin ^2}x$

$\therefore \,\,{\sin ^2}x + {\sin ^4}x = \cos x + {\cos ^2}x = 1$.

एवं $\cos \,2\theta  = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \frac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}$

$\therefore $ $\sin \theta  =  \pm \frac{a}{{\sqrt {{a^2} + {b^2}} }},\,\,\cos \,\theta  =  \pm \frac{b}{{\sqrt {{a^2} + {b^2}} }}$

अब, $\frac{{\sin \,\theta }}{{\cos {\,^8}\theta }} + \frac{{\cos \,\theta }}{{{{\sin }^8}\theta }} $

$= \frac{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)}}{{{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)}^8}}} + \frac{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)}}{{{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)}^8}}}$

 $ = \frac{{a\,{{({a^2} + {b^2})}^4}}}{{{b^8}\,{{({a^2} + {b^2})}^{1/2}}}} + \frac{{b\,{{({a^2} + {b^2})}^4}}}{{{a^8}\,{{({a^2} + {b^2})}^{1/2}}}}$

$ =  \pm \frac{{{{({a^2} + {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\,\left( {\frac{a}{{{b^8}}} + \frac{b}{{{a^8}}}} \right)$.

$ \Rightarrow \,\,p + q = \frac{{\cos \theta }}{{1 + \sin \theta }} + \frac{{2\,\sin \theta }}{{1 + \sin \theta  + \cos \theta }}\,$

$\Rightarrow \,p + q = 1.$

$ = \frac{{1 - \sin \,\phi }}{{\cos \,\phi }}\,.\,\frac{{1 + \cos \,\phi }}{{\sin \,\phi }}$

$ \Rightarrow \,xy + 1 = \frac{{1 - \sin \,\phi + \cos \,\phi - \sin \,\phi \,\cos \,\phi + \sin \phi \cos \phi }}{{\cos \phi \sin \phi }}$

$ = \frac{{1 - \sin \,\phi + \cos \,\phi }}{{\cos \,\phi \sin \,\phi }}$…..$(i)$

$x - y = (\sec \,\phi - \tan \,\phi ) - (\cos ec\,\phi + \cot \,\phi )$

$ = \frac{{1 - \sin \,\phi }}{{\cos \,\phi  }} - \frac{{1 + \cos \,\phi }}{{\sin \,\phi }} = \frac{{\sin \,\phi - {{\sin }^2}\phi - \cos \,\phi - {{\cos }^2}\phi }}{{\cos \,\phi \,\sin \,\phi }}$

$ = \frac{{\sin \,\phi - \cos \,\phi - 1}}{{\cos \,\phi \,\sin \,\phi `}}$…..$(ii)$

$(i)$ व $(ii)$ को जोड़ने पर, $xy + 1 + (x - y) = 0$

$ \Rightarrow x = \frac{{y - 1}}{{y + 1}}$

एवं $2x\,\sec \theta  - y\,\,{\rm{cosec}}\,\theta  = 3$…..$(ii)$

$ \Rightarrow \,\,\frac{{2x}}{{\cos \theta }} - \frac{y}{{\sin \theta }} = 3$

$ \Rightarrow \,\,2x\,\sin \theta  - y\,\cos \theta  - 3\,\sin \theta \cos \theta  = 0$…..$(iii)$

$(i)$ व $(iii)$ को हल करने पर,

$y = \sin \theta $ एवं $x = 2\,\,\cos \theta $

अब, ${x^2} + 4{y^2} = 4\,\,{\cos ^2}\theta  + 4\,\,{\sin ^2}\theta $

$ = 4\,({\cos ^2}\theta  + {\sin ^2}\theta ) = 4$.

$\frac{{1 + \cos y - {{\sin }^2}y}}{{1 + \cos y}} + \frac{{(1 - {{\cos }^2}y) -{{\sin }^2}y}}{{\sin y\,(1 - \cos y)}}$

$ = \frac{{\cos y\,(1 + \cos y)}}{{1 + \cos y}} + 0 = \cos y.$

$ = \frac{{2\,{{\sin }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\,{{\cos }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}$

$ = \frac{{2\,\,\sin \frac{A}{2}\,\left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)}}{{2\,\,\cos \frac{A}{2}\,\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}}$=$\tan \frac{A}{2}$.

ट्रिक :  $A = {60^o}$ रखने पर,

$\frac{{1 + (\sqrt 3 /2) - (1/2)}}{{1 + (\sqrt 3 /2) + (1/2)}} = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

जो कि विकल्प $(c)$ है अर्थात् $\tan \frac{{{{60}^o}}}{2} = \frac{1}{{\sqrt 3 }}$ है।

$ \Rightarrow \,\cos \theta  = (\sqrt 2  + 1)\,\sin \theta \, $

$\Rightarrow \,(\sqrt 2  - 1)\cos \theta  = \sin \theta $

$ \Rightarrow \,\sqrt 2 \,\cos \theta  - \cos \theta  = \sin \theta  $

$\Rightarrow \,\sin \theta  + \cos \theta  = \sqrt 2 \,\cos \theta .$

$\therefore \,\,\,\sec \theta  - \tan \theta  = {e^{ - x}}$…..$(ii)$

$(i)$ व $(ii)$ से,

$\,2\sec \theta  = {e^x} + {e^{ - x}}\,$

$\Rightarrow \,\cos \theta  = \frac{2}{{{e^x} + {e^{ - x}}}}.$

${\cos ^2}A = \sin A\,\tan A = \frac{{{{\sin }^2}A}}{{\cos A}}\,$

$\Rightarrow \,{\cos ^3}A - {\sin ^2}A = 0$

अत: ${\cos ^3}A + {\cos ^2}A = {\sin ^2}A + {\cos ^2}A = 1$

${(m + 2)^2}\,{t^2} + 2(m + 2)\,(2m - 1)t + {(2m - 1)^2} = {(2m + 1)^2}\,(1 + {t^2})$

$ \Rightarrow \,3\,(1 - {m^2})\,{t^2} + (4{m^2} + 6m - 4)\,t - 8m = 0$

$ \Rightarrow \,(3t - 4)\,[(1 - {m^2})\,t + 2m = 0$,

जो कि सत्य है यदि $t = \tan \theta  = \frac{4}{3}$ या $\tan \theta  = \frac{{2m}}{{{m^2} - 1}}$.

$ \Rightarrow {\sin ^2}\theta  = \frac{{{{(a + b)}^2}}}{{4ab}} \le 1 $

$\Rightarrow {(a + b)^2} - 4ab \le 0$

$ \Rightarrow {(a - b)^2} \le 0 $

$\Rightarrow a = b.$

लेकिन $x$ वास्तविक है। अत: ${B^2} - 4AC \ge 0$

$ \Rightarrow {\cos ^2}\theta  \ge 4(1)(1) \Rightarrow {\cos ^2}\theta  \ge 4$, जो कि असम्भव है।

$P=\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}$

Assume $x \leq y \leq z$

for $P$ to be max take $x=0, z=1$

$P =\sqrt{| y |}+\sqrt{| y -1|}+1$

$P =\sqrt{ y }+\sqrt{1- y }+1$

Put $y=\sin ^2 \theta$

$P =\sin \theta+\cos \theta+1$

$P _{\max }=\sqrt{2}+1$

Let $S=\sum \limits_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$

$\Rightarrow \quad S=\frac{1}{\sin 1^{\circ}} \sum \limits_{k=0}^{44} \frac{\sin (k+1-k)^{\circ}}{\cos k^{\bullet} \cos (k+1)^{\circ}}$

$S=\frac{1}{\sin 1^{\circ}}$

$S=\frac{1}{\sin 1^{\circ}} \sum \limits_{k=0}^{44}\left(\tan (k+1)^{\circ}-\tan k^{\circ}\right)$

$S=\frac{1}{\sin 1^{\circ}}\left[\tan 1^{\circ}-\tan 0^{\circ}+\tan 2^{\circ}\right.$

$S=\frac{1}{\sin 1^{\circ}} \tan 45^{\circ}$

$S=\frac{1}{\sin 1^{\circ}}=\frac{1}{0.0174}=57.29$

$\left[\because \sin 1^{\circ}=0.017\right]$

$[S]=[57.29]=57$

We have,

$x(\theta)=|\cos 4 \theta| \cos \theta$ and $y(\theta)=|\cos 4 \theta| \sin \theta$

Given, $C(\theta)=\sum_{n=0}^{\infty} \frac{\cos (n \theta)}{n !}$

$C(0)=\sum_{n=0}^{\infty} \frac{1}{n !}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots=e$

$C(\pi)=\sum_{n=0}^{\infty}(-1)^n \frac{1}{n !} \quad\left[\because \cos n \pi=(-1)^n\right]$

$C(\pi)=1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots=e^{-1}$

(A) $C(0), C(\pi)=e \cdot e^{-1}=1$ T'rue

(B) $C(0)+C(\pi)=e+\frac{1}{e}>2$ True

(C) $C(\theta)>0 \forall \theta \in R$ True

(D) $C^{\prime}(\theta)=\sum_{n=0}^{\infty}-\frac{n \sin (n \theta)}{n !}$

$\therefore \quad C^{\prime}(\theta)=0 \Rightarrow \theta=0$ False

Hence, option $(d)$ is false.

$ 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 $

$ 6 \sin x-4=0 $

$ \sin x=\frac{2}{3} $

${n}=5 \text { (in the given interval) } $

$ x^2+5 x+2=0$

$ x=\frac{-5 \pm \sqrt{17}}{2}$

$ \text { Required interval }(-\infty, 0)$

Required interval $(-\infty, 0)$

$ \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 $

$ \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 $

$ \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 $

$ x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0$

so, sum of real solutions $=-1$

$ a(\sin x-2)=2(\sin x-2)(\sin x+2) $

$ \sin x=2, \quad a=2(\sin x+2) $

$ \Rightarrow a \in[2,6] $

$ p=2 \quad q=6$

$ r=\tan 9^{\circ}+\cot 9^{\circ}-\tan 27-\cot 27 $

$ r=\frac{1}{\sin 9 \cdot \cos 9}-\frac{1}{\sin 27 \cdot \cos 27} $

$=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right] $

$ r=4 $

$ \text { p.q. } r=2 \times 6 \times 4=48$

$2 P \times \sin \frac{\pi}{33}=96 \times 2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$

$2 P \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \sin \frac{\pi}{33}$

$P=3$

$\Rightarrow \tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ}$

$\Rightarrow \frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}$

$\Rightarrow \frac{2 \times 4}{\sqrt{5}-1}-\frac{2 \times 4}{\sqrt{5+1}}$

$\Rightarrow 4$

$\frac{1}{\tan 75^{\circ}}=\cot 75^{\circ}=2-\sqrt{3}$

$\frac{1}{\tan 105^{\circ}}=\cot \left(105^{\circ}\right)=-\cot 75^{\circ}=\sqrt{3}-2$

$\tan 195^{\circ}=\tan 15^{\circ}=2-\sqrt{3}$

$2(2-\sqrt{3})=2 a \Rightarrow a =2-\sqrt{3}$

$a +\frac{1}{ a }=4$

$Ar (\triangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 \tan \theta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(\tan \theta_1+\tan \theta_2\right)$

$BE = AB \left(\tan \theta_1+\cot \theta_1\right)$

$\frac{4}{\sqrt{3}} \tan \theta_1+\cot \theta_1 \Rightarrow \tan \theta_1=\sqrt{3}, \frac{1}{\sqrt{3}}$

$\theta_1=\frac{\pi}{6}$

$\theta_1=\frac{\pi}{3} \quad \theta_2=\frac{\pi}{3}$

$as \frac{\theta_2}{\theta_1} \text { is } \operatorname{largest} \therefore \theta_1=\frac{\pi}{6} \theta_2=\frac{\pi}{3}$

$\therefore x ^2=\frac{(2 \sqrt{3}-3) \times 2}{\tan \theta_1}=\frac{\sqrt{3}(2-\sqrt{3}) \times 2}{\tan \frac{\pi}{6}}$

$x ^2=12-6 \sqrt{3}=(3-\sqrt{3})^2$

$x =3-\sqrt{3}$

Perimeter of $\triangle C E D$

$= CD + DE + CE$

$=3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) \times 2=6$

$\tan (\alpha+\beta)=\frac{1-\frac{4}{3}}{1+\frac{4}{3} \times 1}=\frac{-1}{7}$

$=2 \sin \left(\frac{11 \pi-10 \pi}{22}\right) \sin \left(\frac{11 \pi-8 \pi}{22}\right) \sin \left(\frac{11 \pi-6 \pi}{22}\right)$ $\sin \left(\frac{11 \pi-4 \pi}{22}\right) \sin \left(\frac{11 \pi-2 \pi}{22}\right)$

$=2 \cos \frac{\pi}{11} \cos \frac{2 \pi}{11} \cos \frac{3 \pi}{11} \cos \frac{4 \pi}{11} \cos \frac{5 \pi}{11}$

$=\frac{2 \sin \frac{32 \pi}{11}}{2^{5} \sin \frac{\pi}{11}}$

$=\frac{1}{16}$

$=\frac{\sin \left(3 \times \frac{\pi}{7}\right)}{\sin \frac{\pi}{7}} \times \cos \left(\frac{\frac{2 \pi}{7}+\frac{6 \pi}{7}}{2}\right)$

$=\frac{2 \sin \left(\frac{3 \pi}{7}\right)}{2 \sin \frac{\pi}{7}} \times \cos \left(\frac{4 \pi}{7}\right)$

$=\frac{\sin \left(\frac{7 \pi}{7}\right)+\sin \left(\frac{-\pi}{7}\right)}{2 \sin \frac{\pi}{7}}$

$=\frac{-\sin \frac{\pi}{7}}{2 \sin \frac{\pi}{7}}$

$=-\frac{1}{2}$

$=16 \sin 40^{\circ} \sin 20^{\circ} \sin 80^{\circ}$

$=4(4 \sin (60-20) \sin (20) \sin (60+20))$

$=4 \times \sin \left(3 \times 20^{\circ}\right)$

${[\because \sin 3 \theta=4 \sin (60-\theta) \times \sin \theta \times \sin (60+\theta)]}$

$=4 \times \sin 60^{\circ}$

$=4 \times \frac{\sqrt{3}}{2}=2 \sqrt{3}$

$=\sin 12^{\circ}-2 \cos 42^{\circ} \sin 30^{\circ}$

$=\sin 12^{\circ}-\sin 48^{\circ}$

$=-2 \cos 30^{\circ} \sin 18^{\circ}$

$=-2 \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{5}-1}{4}$

$=\frac{\sqrt{3}}{4}(1-\sqrt{5})$

$\sin 10^{\circ} \frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \cdot \frac{1}{4} \sin 30^{\circ}$

$\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \sin 10^{\circ}\left(\cos 20^{\circ}-\frac{1}{2}\right)$

$=\frac{1}{32}\left(2 \sin 10^{\circ} \cos 20^{\circ}-\sin 10^{\circ}\right)$

$=\frac{1}{32}\left(\sin 30^{\circ}-\sin 10^{\circ}-\sin 10^{\circ}\right)$

$=\frac{1}{32}\left(\frac{1}{2}-2 \sin 10^{\circ}\right)$

$=\frac{1}{64}\left(1-4 \sin 10^{\circ}\right)$

$=\frac{1}{64}-\frac{1}{16} \sin 10^{\circ}$

Hence $\alpha=\frac{1}{64}$

$16+\alpha^{-1}=80$

$\Rightarrow k +1 \in\left[-\sqrt{3^{2}+4^{2}}, \sqrt{3^{2}+4^{2}}\right]$

$\Rightarrow k +1 \in[-5,5]$

$\Rightarrow k \in[-6,4]$

No. of integral values of $k =11$

$16\left[4 \sin 2 \theta \cos ^{2} 2 \theta+2 \cos ^{2} 2 \theta-1\right]$

Now:

$\sin \theta+\cos \theta=\frac{1}{2}$

$1+\sin 2 \theta=\frac{1}{4}$

$\sin 2 \theta=-\frac{3}{4}$

$\cos ^{2} 2 \theta=1-\frac{9}{16}=\frac{7}{16}$

$16\left[-4(-3 / 4) \times \frac{7}{16}+2 \times \frac{7}{16}-1\right]$

$16\left[\frac{-7}{16}-1\right] \Rightarrow-23$

$2 \sin ^{2} \frac{\pi}{8} \sin ^{2} \frac{2 \pi}{8} \sin ^{2} \frac{3 \pi}{8}$

$\sin ^{2} \frac{\pi}{8} \sin ^{2} \frac{3 \pi}{8}$

$\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8}$

$\frac{1}{4} \sin ^{2}\left(\frac{\pi}{4}\right)=\frac{1}{8}$

Let $\operatorname{cosec} 18^{\circ}=\mathrm{x}=\sqrt{5}+1$

$\Rightarrow \mathrm{x}-1=\sqrt{5}$

Squaring both sides, we get

$x^{2}-2 x+1=5$

$\Rightarrow x^{2}-2 x-4=0$

$\log _{10} \sin x+\log _{10} \cos x=-1$

$\Rightarrow \quad \log _{10} \sin x \cdot \cos x=-1$

$\Rightarrow \quad \sin x \cdot \cos x=\frac{1}{10}$ $....(1)$

$\log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10} n-1\right)$

$\Rightarrow \quad \sin x+\cos x=10^{\left(\log _{10} \sqrt{n}-\frac{1}{2}\right)}=\sqrt{\frac{n}{10}}$

by squaring

$1+2 \sin x \cdot \cos x=\frac{n}{10}$

$\Rightarrow \quad 1+\frac{1}{5}=\frac{ n }{10} \quad \Rightarrow \quad n =12$

$\cos ^{2}\left(\frac{x+y}{2}\right)-\cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)$

$+\frac{1}{4} \cdot \cos ^{2}\left(\frac{x-y}{2}\right)+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow\left(\cos \left(\frac{x+y}{2}\right)-\frac{1}{2} \cos \left(\frac{x-y}{2}\right)\right)^{2}+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x-y}{2}\right)=0$ and $\cos \left(\frac{x+y}{2}\right)=\frac{1}{2} \cos \left(\frac{x-y}{2}\right)$

$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$

$\therefore \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

put, $\theta=\frac{\pi}{24}$

$\left\{\therefore \cos \frac{\pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \, \& \, \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\right\}$

$\Rightarrow \cot \left(\frac{\pi}{24}\right)=\frac{1+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)}$

$=\frac{(2 \sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$

$=\frac{2 \sqrt{6}+2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$

$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$

$\tan \alpha . \tan \beta=\frac{\mathrm{k}-1}{\mathrm{k}+1}$

$\tan (\alpha+\beta)=\frac{\frac{\lambda \sqrt{2}}{\mathrm{k}+1}}{1-\frac{\mathrm{k}-1}{\mathrm{k}+1}}=\frac{\lambda \sqrt{2}}{2}=\frac{\lambda}{\sqrt{2}}$

$\Rightarrow \frac{\lambda^{2}}{2}=50 \Rightarrow \lambda=10 \;and\;-10$

$=\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}=\frac{1}{2} \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}}$

$\sin \beta=\frac{1}{\sqrt{10}} \Rightarrow \tan \beta=\frac{1}{3} \Rightarrow \tan 2 \beta=\frac{3}{4}$

$\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}=1$

If $\cos \,(\,\alpha  + \,\beta )\, = \,\frac{3}{5}$ then $\tan \,(\,\alpha  + \,\beta )\, = \,\frac{3}{4}$ and if $\sin \,(\,\alpha  - \,\beta )\, = \,\frac{5}{{13}}$ then $\tan \,(\,\alpha  - \,\beta )\, = \,\frac{5}{{12}}$

(since $\alpha  - \,\beta $ here lies in the first quadrant)

Now $\tan \,(\,2\alpha ) = \tan \{ (\alpha \, + \,\beta ) + (\alpha \, - \,\beta )\} $

$ = \,\frac{{\tan \,(\alpha \, + \,\beta ) + \tan \,(\alpha \, - \,\beta )}}{{1 - \tan \,(\alpha \, + \,\beta ).\tan \,(\alpha \, - \,\beta )}}$ 

$ = \frac{{\frac{4}{3} + \frac{5}{{12}}}}{{1 - \frac{4}{3}.\frac{5}{{12}}}} = \frac{{63}}{{16}}$

${F_6}(x)\, = \,\frac{{{{\sin }^6}x + {{\cos }^6}x}}{6} = $ $\frac{{1 - 3{{\sin }^2}x.{{\cos }^2}x\,({{\sin }^2}x\, + \,{{\cos }^2}x)}}{6}$

$ = \frac{1}{6} - \frac{1}{2}{\sin ^2}x.{\cos ^2}x$

${F_4}\left( x \right) - {F_6}(x) = \frac{1}{4} - \frac{1}{6} = \frac{{6 - 4}}{{24}} = \frac{2}{{24}} = \frac{1}{{12}}$

$ = \,5\left( {\sin \,\theta \,\cos \frac{\pi }{6}\, - \,\cos \,\theta \sin \frac{\pi }{6}} \right)\, + \,3\cos \,\theta $

$ = \frac{{5\sqrt 3 }}{2}\sin \,\theta \, + \,\frac{1}{2}\cos \,\theta $

Maximum value is  $\sqrt {{{\left( {\frac{{5\sqrt 3 }}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} \, = \,\sqrt {\frac{{76}}{4}} \, = \,\sqrt {19} $

$ = \,\,\sin \,{10^o}\,\sin \,{30^o}\,\sin \,{50^o}\,\sin \,{70^o}$

$ = \,\,\sin \,{30^o}\,\{ \sin \,{10^o}\sin \,({60^o} - {10^o})\,\sin ({60^o} + {10^o})\,\} $

$ = \,\,\sin \,{30^o}\,\left\{ {\frac{1}{4}\sin \,3({{10}^o})} \right\}$

$ = \,\frac{1}{2}\left( {\frac{1}{4} \times \frac{1}{2}} \right)$

$ = \,\frac{1}{{16}}$

$ \Rightarrow \frac{1}{2}\,(1 + \,\cos \,{20^o} - (\cos \,{60^o} + \cos \,{40^o})\, + 1 + \,\,\cos {100^o})$

$ \Rightarrow \frac{1}{2}\,\left( {\frac{3}{2} + \,\cos \,{{20}^o} + 2\sin \,{{70}^o}\sin \,( - {{30}^o})} \right)$

$ \Rightarrow \frac{1}{2}\,\left( {\frac{3}{2} + \,\cos \,{{20}^o} - \sin \,{{70}^o}} \right)$

$ \Rightarrow \frac{3}{4}$

Let $\,{\cos ^2}2\theta \, =t $

$ \Rightarrow \,1\, - \,\,{\cos ^2}2\theta \, + \,{\cos ^4}2\theta \, = \frac{3}{4}$

$ \Rightarrow t = \frac{1}{2}\, \Rightarrow \,\,{\cos ^2}2\theta \, = \frac{1}{2}\,$

$ \Rightarrow 2\,{\cos ^2}2\theta  - 1 = 0\, \Rightarrow \,\cos 4\theta \, = 0$

$ \Rightarrow \,4\theta \, = (2n + 1)\frac{\pi }{2}$

$ \Rightarrow \,\theta \, = (2n + 1)\frac{\pi }{8} \Rightarrow \theta  = \frac{\pi }{8},\frac{{3\pi }}{8} \in \left[ {0,\frac{\pi }{2}} \right]$

Sum of values of $\theta $ is $\frac {\pi }{2}$

$ = 3\,(1 - 2\sin \,2\theta  + {\sin ^2}2\theta ) + \,6 + 6\sin 2\theta  + \,4\,{\sin ^6}\theta $

$ = \,9 + 3{\sin ^2}2\theta  + 4\,{\sin ^6}\theta $

$ = \,9 + 12{\sin ^2}\theta {\cos ^2}\theta  + 4\,{(1 - {\cos ^2}\theta )^3}$

$ = \,9 + 12(1 - {\cos ^2}\theta ){\cos ^2}\theta  + 4\,(1 - 3{\cos ^2}\theta  + 3{\cos ^4}\theta  - {\cos ^6}\theta )$

$ = \,13 + 12{\cos ^2}\theta  - 12{\cos ^4}\theta  - 12{\cos ^2}\theta  + 12{\cos ^4}\theta  - 4{\cos ^6}\theta $

$ = \,13 - 4{\cos ^6}\theta $

$\tan A+\tan B=\frac{10}{3}$

$\tan A \times \tan B=\frac{-25}{3}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\frac{\frac{10}{3}}{1+\frac{25}{3}}$

$\frac{\frac{10}{3}}{\frac{28}{3}}=\frac{10}{28}=\frac{5}{14}$

$\therefore \tan (A+B)=\frac{5}{14}$

$\therefore \sin (A+B)=\frac{5}{\sqrt{221}}$

$\cos (A+B)=\frac{14}{\sqrt{221}}$

$\therefore 3 \sin ^2(A+B)-10 \sin (A+B) \cos (A+B)-25 \cos ^2(A+B)$

$=3 \times \frac{25}{221}-10 \times \frac{70}{221}-25 \times \frac{196}{221}$

$=\frac{75-700-4900}{221}$

$=-25$

$5\,{\tan ^2}x\, - 5{\cos ^2}x = 2(2{\cos ^2}x - 1) + 9$

$ \Rightarrow \,5\,{\tan ^2}x\, - 5{\cos ^2}x = 4{\cos ^2}x - 2 + 9$

$ \Rightarrow \,5\,{\tan ^2}x = 9{\cos ^2}x + 7$

$ \Rightarrow \,5\,({\sec ^2}x - 1) = 9{\cos ^2}x + 7$

Let ${\cos ^2}x = t$

$ \Rightarrow \frac{5}{t} - 9t - 12 = 0$

$ \Rightarrow 9{t^2} + 12t - 5 = 0$

$ \Rightarrow 9{t^2} + 15t - 3t - 5 = 0$

$ \Rightarrow (3t - 1)(3t + 5) = 0$

$ \Rightarrow t\, = \frac{1}{3}$ as $t\, \ne  - \frac{5}{3}$

$\cos \,2x = 2{\cos ^2}x - 1 = 2\left( {\frac{1}{3}} \right) - 1 =  - \frac{1}{3}$

$\cos \,4x = 2{\cos ^2}2x - 1 = 2{\left( { - \frac{1}{3}} \right)^2} - 1 =  - \frac{7}{9}$

$ \Rightarrow \,\frac{1}{{\sqrt 3 }}\, = \,\frac{y}{{1 - \tan \,A\,\tan \,B}}$ where $y\, = \tan \,A\, + \tan \,B$

$ \Rightarrow \tan \,A\,\tan \,B\, = \,1 - \sqrt 3 y$ Also $AM \geqslant GM$

$ \Rightarrow \,\frac{{\tan \,A\, + \,\tan \,B}}{2}\, \geqslant \,\sqrt {\tan \,A\,\tan \,B} $

$ \Rightarrow \,y\, \geqslant \,2\sqrt {1 - \sqrt 3 y} $

$ \Rightarrow \,{y^2}\, \geqslant \,4 - 4\sqrt 3 y$

$ \Rightarrow \,{y^2}\, + \,4\sqrt 3 y - 4 \geqslant 0$

$ \Rightarrow \,y\, \leqslant \, - \,2\sqrt 3  - 4$ or $ \Rightarrow \,y\,  \geqslant  \, - \,2\sqrt 3  + 4$

( $y\, \leqslant \, - \,2\sqrt 3  - 4$ is not possible as $\tan A\,\tan B\, > 0$ 

$4\, + \,2(1 - {\cos ^2}x){\cos ^2}x - 2{\cos ^4}x$

$ - 4\left\{ {{{\cos }^4}x - \frac{{{{\cos }^2}x}}{2} - 1 + \frac{1}{{16}} - \frac{1}{{16}}} \right\}\, - \,4\left\{ {{{\left( {{{\cos }^2}x - \frac{1}{4}} \right)}^2} - \frac{{17}}{{16}}} \right\}$

$0\, \leqslant \,{\cos ^2}x\, \leqslant 1$

$ - \frac{1}{4}\, \leqslant \,\,{\cos ^2}x\, - \,\frac{1}{4}\, \leqslant \,\frac{3}{4}$

$0\, \leqslant \,{\left( {\,{{\cos }^2}x - \frac{1}{4}} \right)^2}\, \leqslant \,\frac{9}{{16}}$

$ - \frac{{17}}{{16}}\, \leqslant \,{\left( {\,{{\cos }^2}x - \frac{1}{4}} \right)^2}\, - \frac{{17}}{{16}}\, \leqslant \,\frac{9}{{16}}\, \leqslant \, - \frac{{17}}{{16}}\,$

$\frac{{17}}{4} \geqslant  - \,4\left\{ {\,{{\left( {\,{{\cos }^2}x - \frac{1}{4}} \right)}^2} - \frac{{17}}{{16}}\,\,} \right\}\, \geqslant \,\frac{1}{2}$

$M = \frac{{17}}{4}$

$m = \frac{1}{2}$

$M - m\, = \,\frac{{17}}{4} - \frac{2}{4}\, = \frac{{15}}{4}$

$f(x)\, = \,3\,\sin x + 4\,\cos x$

Now,

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \,3\,\sin \left( {\frac{{ - 11\pi }}{6}} \right)\, + 4\,\cos \left( {\frac{{ - 11\pi }}{6}} \right)\,$

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \,3\,\sin \left( { - 2\pi  + \frac{\pi }{6}} \right)\, + 4\,\cos \left( { - 2\pi  + \frac{\pi }{6}} \right)\,$

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \,3\,\sin \left\{ { - \left( {2\pi  - \frac{\pi }{6}} \right)} \right\}\, + 4\,\cos \left\{ { - \left( {2\pi  - \frac{\pi }{6}} \right)} \right\}$

$\left\{ {For\,\,odd\,\,functions\,\begin{array}{*{20}{c}}
  {\sin \theta \, = \, - \,\sin \theta } \\ 
  {and\,\,\cos \,( - \theta ) =  - \cos \,\,\theta } 
\end{array}} \right\}$

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \, - 3\,\sin \left( {2\pi  - \frac{\pi }{6}} \right)\, - 4\,\cos \left( {2\pi  - \frac{\pi }{6}} \right)$

$ \Rightarrow \,f\left( {\frac{{ - 11\,\pi }}{6}} \right)\, = \,\, + 3\sin \left( {\frac{\pi }{6}} \right) - 4\,\cos \left( {\frac{\pi }{6}} \right)$

$ \Rightarrow \,f\left( {\frac{{ - 11\,\pi }}{6}} \right)\, = \,\,3 \times \frac{1}{2} - 4\, \times \frac{{\sqrt 3 }}{2}$

or $\,f\left( {\frac{{ - 11\,\pi }}{6}} \right)\, = \,\,\frac{3}{2} - 2\sqrt 3 $

$ = \frac{1}{4}(1 - 2{\sin ^2}{\cos ^2}x) - \frac{1}{6}(1 - 3{\sin ^2}{\cos ^2}x)$

$ = \frac{1}{4} - \frac{1}{6} = \frac{{3 - 2}}{{12}} = \frac{1}{{12}}$

$ = \frac{1}{{\sin A - \cos A}}\left\{ {\frac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A}}} \right\}$

$ = \frac{{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}A}}{{\sin A\cos A}}$ $ = 1 + \sec A\cos ecA$

$\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$

$=\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x\right\}$

$=\cos \left(6 x-\frac{11 x}{2}\right)+\sin \left(6 x-\frac{11 x}{2}\right)$

$=\cos \frac{x}{2}+\sin \frac{x}{2}$

$=\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}}$

$=\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow \text { Option (B) is correct. }$

$\cot x=-\frac{5}{\sqrt{11}}$

$\frac{1-\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}}=-\frac{5}{\sqrt{11}}$

$\tan \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}}$

$\tan \frac{x}{2}=\sqrt{11}$, As $\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

$\tan \frac{x}{2}+\tan \frac{z}{2}=\frac{2 y}{x+y+z}$

$\frac{\Delta}{S(S-x)}+\frac{\Delta}{S(S-z)}=\frac{2 y}{2 S}$

$\frac{\Delta}{ S }\left(\frac{2 S -( x + z )}{( S - x )( S - z )}\right)=\frac{ y }{ S }$

$\Rightarrow \frac{\Delta y}{S(S-x)(S-z)}=\frac{y}{S}$

$\Rightarrow \quad \Delta^2=(S-x)^2(S-z)^2$

$\Rightarrow \quad S ( S - y )=( S - x )( S - z )$

$\Rightarrow \quad(x+y+z)(x+z-y)=(y+z-x)(x+y-z)$

$\Rightarrow \quad(x+z)^2-y^2=y^2-(z-x)^2$

$\Rightarrow \quad(x+z)^2+(x-z)^2=2 y^2$

$\Rightarrow \quad x ^2+ z ^2= y ^2 \Rightarrow \angle Y =\frac{\pi}{2}$

$\Rightarrow \quad \angle Y =\angle X +\angle Z$

$\tan \frac{x}{2}=\frac{\Delta}{S(S-x)}$

$\tan \frac{x}{2}=\frac{\frac{1}{2} x z}{\frac{(y+z)^2-x^2}{4}}$

(IMAGE)

$\tan \frac{x}{2}=\frac{2 x z}{y^2+z^2+2 y z-x^2}$

$\tan \frac{x}{2}=\frac{2 x z}{2 z^2+2 y z} \quad \text { (using } y^2=x^2+z^2 \text { ) }$

$\tan \frac{x}{2}=\frac{x}{y+z}$

$f(\theta)  =\sin ^2 2 \theta-\sin 2 \theta+2$

$f^{\prime}(\theta)  =2(\sin 2 \theta) \cdot(2 \cos 2 \theta)-2 \cos 2 \theta$

$ =2 \cos 2 \theta(2 \sin 2 \theta-1)$

critical points

so, minimum at $\theta=\frac{\pi}{12}, \frac{5 \pi}{12}$

$\lambda_1+\lambda_2=\frac{1}{12}+\frac{5}{12}=\frac{6}{12}=\frac{1}{2}$

$X :\{ x : f ( x )=0\}$

$f(x)=0 \Rightarrow \sin (x \cos x)=0 \Rightarrow \cos x=n \Rightarrow \cos x=1,-1,0 \Rightarrow x=\frac{n \pi}{2}$

$x =\left\{\frac{ n \pi }{2}: \pi \in N \right\}-\left\{\frac{\pi}{2}, \pi \cdot \frac{3 \pi}{2}, 2 \pi,\right\}$

$g(x)=\cos (2 \pi \sin x)$

$Z=\{x: g(x)=0\}$

$\cos (2 \pi \sin x)=0 \Rightarrow 2 \pi \sin x=(2 n+1) \frac{\pi}{2} \Rightarrow \sin x-\frac{(2 n+1)}{4}$

$\sin x=-\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{3}{4}$

$Z=\left\{n \pi \pm \sin ^{-1}\left(\frac{1}{4}\right), m \pi \pm \sin ^{-1}\left(\frac{3}{4}\right), n \in I\right\}$

$Y =\{ x : f ( x )=0\}$

$f(x)=\sin (\pi \cos x) \Rightarrow f^{\prime}(x)=\cos (\pi \cos x) \cdot(-\pi \sin x)=0$

$\sin x=0 \Rightarrow x=m \pi \text {. }$

$\cos (\pi \cos x)=0 \Rightarrow \pi \cos x=(2 n+1) \frac{\pi}{2} \Rightarrow \cos x-\frac{(2 n+1)}{2} \Rightarrow \cos x=-\frac{1}{2} \cdot \frac{1}{2}$

$Y=\left\{2 \pi, n \pi \pm \frac{\pi}{3}\right\}-\left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}, 2 \pi, \ldots . .\right\}$

$W=\left\{x: g^{\prime}(x)=0\right\}$

$g(x)=\cos (2 \pi \sin x) \Rightarrow g^{\prime}(x)=-\sin (2 \pi \sin x) \cdot(2 \pi \cos x)=0$

$\cos x =0 \Rightarrow x =(2 n +1) \frac{\pi}{2}$

$\sin (2 \pi \sin x)=0 \Rightarrow 2 \pi \sin x=n \pi \Rightarrow \sin x=\frac{\pi}{2}=-1-\frac{1}{2} \cdot 0 \cdot \frac{1}{2} \cdot 1$

$W=\left\{\frac{n \pi}{2}, n \pi \pm \frac{\pi}{6}, \pi \in I\right\}-\left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \pi, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \ldots\right\}$

Now check the options

$\sqrt{3} \cos x+\frac{2 b}{a} \sin x=\frac{c}{a}$

As $\alpha, \beta$ are roots $\Rightarrow \sqrt{3} \cos \alpha+\frac{2 b }{ a } \sin \alpha=\frac{ c }{ a }$    $. . . . . .(1)$

$(1)$ $-(2)$

$\sqrt{3} \cos \beta+\frac{2 b}{a} \sin \beta=\frac{c}{a}$   $. . . . . .(2)$

$\sqrt{3}(\cos \alpha-\cos \beta)+\frac{2 b}{a}(\sin \alpha-\sin \beta)=0$

$\sqrt{3}\left(2 \sin \frac{\alpha+\beta}{2} \sin \frac{\beta-\alpha}{2}\right)+\frac{2 b}{a}\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)=0$

Given $\alpha+\beta=\frac{\pi}{3}$

$\sqrt{3}\left(2 \cdot \frac{1}{2} \sin \frac{\beta-\alpha}{2}\right)+2 \frac{2 b}{a}\left(\frac{\sqrt{3}}{2} \sin \frac{\alpha-\beta}{2}\right)=0$

$\Rightarrow 1=\frac{2 b}{a} \quad \Rightarrow \frac{b}{a}=\frac{1}{2}$

$\Rightarrow \sqrt{3} \sec x+\csc x=2(\cot x-\tan x)$

$\Rightarrow \frac{(\sqrt{3} \sec x+\csc x)}{2}=\cot x-\tan x$

$\Rightarrow \frac{\sqrt{3}}{2} \sec x +\frac{1}{2} \csc x =\cot x -\tan x$ by dividing both sides by $2$

We know that $\sec x=\frac{1}{\cos x}, \csc x=\frac{1}{\sin x}, \tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$ $\Rightarrow \frac{\sqrt{3} 1}{2 \cos x}+\frac{11}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$

$\Rightarrow \frac{\sqrt{3} \sin x}{2 \sin x \cos x}+\frac{1 \cos x}{2 \sin x \cos x}=\frac{\cos ^2 x}{\sin x \cos x}-\frac{\sin ^2 x}{\cos x \sin x}$

$\Rightarrow \frac{\sqrt{3}}{2} \sin x +\frac{1}{2} \cos x =\cos ^2 x -\sin ^2 x$

$\Rightarrow \sin \frac{\pi}{3} \sin x+\cos \frac{\pi}{3} \cos x=\cos 2 x$ where $\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}$ and $\frac{1}{2}=\cos \frac{\pi}{3}$

$\Rightarrow \cos \left(\frac{\pi}{3}-x\right)=\cos 2 x$

$\Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right)$ since if $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha$

$\Rightarrow 2 x =2 n \pi+ x -\frac{\pi}{3}$ or $2 x =2 n \pi- x +\frac{\pi}{3}$

$\Rightarrow x=2 n \pi-\frac{\pi}{3}$ or $3 x=2 n \pi+\frac{\pi}{3}$ or $x=\frac{2 n \pi}{3}+\frac{\pi}{9}$

For $n =0, x =\frac{-\pi}{3}, \frac{\pi}{9}$

For $n =-1, x =\frac{-7 \pi}{3}$ which does not lie between $(-\pi, \pi)$

and $x=\frac{-2 \pi}{3}+\frac{\pi}{9}=\frac{-5 \pi}{9} \in(-\pi, \pi)$

For $n =2, x =\frac{5 \pi}{3}$ does not lie between $(-\pi, \pi)$

and $x=\frac{2 \pi}{3}+\frac{\pi}{9}=\frac{7 \pi}{9} \in(-\pi, \pi)$

$\therefore x =\frac{-\pi}{3}, \frac{\pi}{9}, \frac{-5 \pi}{9}, \frac{7 \pi}{9}$

Sum of all the solutions $=\frac{-\pi}{3}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=\frac{-3 \pi}{9}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=0$

$\cos (\alpha  - \beta ) = 1$

$\Rightarrow$  $\alpha  - \beta  = 0$

$\Rightarrow$  $\alpha  = \beta $

$\cos 2\alpha  = \frac{1}{e}$

एवं $ - 2\pi  < 2\alpha  < 2\pi $

अत: इसके चार हल हैं।

दो पदों को एक साथ लेने पर, ${f_n}(\theta ) = \frac{{\sin \theta }}{{\cos \theta }}\left[ {\frac{{2{{\cos }^2}\theta }}{{\cos 2\theta }}.\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}.....} \right]$

पुन: प्रथम दो पदों को समायोजित करने पर,

${f_n}(\theta ) = \tan 2\theta \left[ {\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}.....} \right] = \tan ({2^n}\theta )$

$\therefore {f_2}\left( {\frac{\pi }{{16}}} \right) = \tan \frac{{4\pi }}{{16}} = \tan \left( {\frac{\pi }{4}} \right) = 1$

${f_3}\left( {\frac{\pi }{{32}}} \right) = \tan \frac{{8\pi }}{{32}} = \tan \left( {\frac{\pi }{4}} \right) = 1$

${f_4}\left( {\frac{\pi }{{64}}} \right) = \tan \frac{{16\pi }}{{64}} = \tan \,\left( {\frac{\pi }{4}} \right) = 1$
${f_5}\left( {\frac{\pi }{{128}}} \right) = \tan 32\frac{\pi }{{128}} = \tan \,\left( {\frac{\pi }{4}} \right) = 1$.

$\cos {15^o} = \cos ({45^o} - {30^o}) = \frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}=$ अपरिमेय

$\therefore \,\,\,\sin {15^o}\cos {15^o} = \frac{1}{2}(2\sin {15^o}\cos {15^o})$

$ = \frac{1}{2}\sin {30^o} = \frac{1}{2}.\frac{1}{2} = \frac{1}{4} =$ परिमेय

$\therefore \, \sin {15^o}\cos {75^o} = \sin {15^o}\sin {15^o} = {\sin ^2}{15^o}$

$ = {\left( {\frac{{\sqrt 3  - 1}}{{2\sqrt 2 }}} \right)^2} = \frac{{4 - 2\sqrt 3 }}{8}=$ अपरिमेय

$\Rightarrow$  $\frac{1}{{\cos (\theta  - \alpha )}},\frac{1}{{\cos \theta }},\frac{1}{{\cos (\theta  + \alpha )}}$ स.श्रे. में होंगे

अत: $\frac{2}{{\cos \theta }} = \frac{1}{{\cos (\theta  - \alpha )}} + \frac{1}{{\cos (\theta  + \alpha )}}$

$ = \frac{{\cos (\alpha  + \theta ) + \cos (\theta  - \alpha )}}{{{{\cos }^2}\theta  - {{\sin }^2}\alpha }}$

$\Rightarrow$  $\frac{2}{{\cos \theta }} = \frac{{2\cos \theta \cos \alpha }}{{{{\cos }^2}\theta  - {{\sin }^2}\alpha }}$

$\Rightarrow$  ${\cos ^2}\theta  - {\sin ^2}\alpha  = {\cos ^2}\theta \cos \alpha $

$\Rightarrow$ ${\cos ^2}\theta \,(1 - \cos \alpha ) = {\sin ^2}\alpha $

$\Rightarrow$  ${\cos ^2}\theta \left( {2{{\sin }^2}\frac{\alpha }{2}} \right) $

$= 4{\sin ^2}\frac{\alpha }{2}{\cos ^2}\frac{\alpha }{2}$

${\cos ^2}\theta {\sec ^2}\frac{\alpha }{2} = 2$

$\Rightarrow \cos \theta \sec \frac{\alpha }{2} =  \pm \sqrt 2 $.

जब $\alpha  + \beta  + \gamma  = \pi $.

$\therefore $ $\sin \alpha  + \sin \beta  + \sin \gamma  > 0$

अत: $\sin \alpha  + \sin \beta  + \sin \gamma $ का न्यूनतम मान हमेशा धनात्मक होता है।

$ = \frac{{{{(\cos x - \sin x)}^2}}}{{({{\cos }^2}x - {{\sin }^2}x)}} $

$= \frac{{\cos x - \sin x}}{{\cos x + \sin x}} = \frac{{1 - \tan x}}{{1 + \tan x}}$

$ = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \left( {\frac{\pi }{4}} \right)\sin x}} = \tan \left( {\frac{\pi }{4} - x} \right)$.

$y = 1 + {\sin ^2}\phi + {\sin ^4}\phi + .... = \frac{1}{{(1 - {{\sin }^2}\phi )}} = \frac{1}{{{{\cos }^2}\phi }}$

$z = 1 + {\cos ^2}\phi {\sin ^2}\phi + {\cos ^4}\phi {\sin ^4}\phi + .. = \frac{1}{{(1 - {{\cos }^2}\phi {{\sin }^2}\phi )}}$

अब $xyz = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi (1 - {{\cos }^2}\phi {{\sin }^2}\phi )}}$

$xy + z = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi }} + \frac{1}{{1 - {{\cos }^2}\phi {{\sin }^2}\phi }}$

$ = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi (1 - {{\cos }^2}\phi {{\sin }^2}\phi )}} = xyz$

जो कि $(b)$ में दिया है।

एवं $x + y + z = xyz$, जो $(c)$ में दिया है .

अत: $x = \frac{\pi }{{12}}$पर अधिकतम मान होगा।

$y = \frac{{1 - 3{{\tan }^2}x}}{{3 - {{\tan }^2}x}} = \frac{{\frac{1}{3} - {{\tan }^2}x}}{{1 - \frac{1}{3}.{{\tan }^2}x}}$

अत: $y$ के परिभाषित होने के लिए इसे $\frac{1}{3}$ एवं $3$ के बीच में स्थित नहीं होना चाहिए।

$ = \sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}} \times 1$ 

$ \times \sin \left( {\pi - \frac{{5\pi }}{{14}}} \right)\sin \left( {\pi - \frac{{3\pi }}{{14}}} \right)\sin \left( {\pi - \frac{\pi }{{14}}} \right)$ 

$ = {\left[ {\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}} \right]^2} = \frac{1}{{64}}$.

$ = \frac{{\sqrt 3 \cos 20^\circ - \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }} $

$= \frac{{2\left[ {\frac{{\sqrt 3 }}{2}\cos 20^\circ - \frac{1}{2}\sin \,20^\circ } \right]}}{{\frac{2}{2}\sin 20^\circ \cos 20^\circ }}$

$ = \frac{{4\cos (20^\circ + 30^\circ )}}{{\sin 40^\circ }} $

$= \frac{{4\cos 50^\circ }}{{\sin 40^\circ }} = \frac{{4\sin 40^\circ }}{{\sin 40^\circ }} = 4$.

$ = \tan \alpha + 2\tan \,2\alpha + 4\left[ {\frac{{\sin 4\alpha }}{{\cos 4\alpha }} + 2\frac{{\cos \,8\alpha }}{{\sin \,8\alpha }}} \right]$

$ = \tan \alpha + 2\tan 2\alpha + $

$4\left[ {\frac{{\cos \,4\alpha \,\cos \,8\alpha + \sin \,4\alpha \,\sin \,8\alpha + \cos \,4\alpha \cos \,8\alpha }}{{\sin \,8\alpha \,\cos \,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\tan \,2\alpha + 4\left[ {\frac{{\cos \,4\alpha + \cos \,4\alpha \,\cos \,8\alpha }}{{\sin \,8\alpha \cos \,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\,\tan \,2\alpha + 4\,\left[ {\frac{{\cos \,\,4a(1 + \cos \,8\alpha )}}{{\cos \,4\alpha \sin \,8\alpha }}} \right]$

$ = \tan \alpha + 2\tan \,2\alpha + 4\left[ {\frac{{2{{\cos }^2}4\alpha }}{{2\sin \,4\alpha \,\,\cos \,\,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\tan \,2\alpha + 4\cot \,4\alpha $$ = \tan \alpha + 2(\tan 2\alpha + 2\cot 4\alpha )$

$ = \tan \,\alpha + 2\left[ {\frac{{\sin \,\,2\alpha }}{{\cos 2\alpha }} + 2\frac{{\cos \,4\alpha }}{{\sin \,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\left[ {\frac{{\cos \,2\alpha (1 + \cos \,4\alpha )}}{{\sin \,4\alpha \cos \,2\alpha }}} \right]$

$ = \tan \alpha + 2\cot 2\alpha = \frac{{\sin \,\alpha }}{{\cos \,\alpha }} + \frac{{2\cos \,2\alpha }}{{\sin \,2\alpha }}$

$ = \frac{{\cos \,\alpha + \cos \alpha \cos \,2\alpha }}{{\sin \,2\alpha \cos \alpha }}$

$ = \frac{{1 + \cos \,2\alpha }}{{\sin \,2\alpha }} = \frac{{2{{\cos }^2}\alpha }}{{2\sin \alpha \cos \alpha }} = \cot \,\alpha $.

$ = \frac{{\sin \,{2^4}\frac{{2\pi }}{{15}}}}{{{2^4}\sin \frac{{2\pi }}{{15}}}} $

$= \frac{{\sin \,\frac{{32\pi }}{{15}}}}{{16\,\sin \frac{{2\pi }}{{15}}}} $

$= \frac{1}{{16}}\frac{{\sin \frac{{2\pi }}{{15}}}}{{\sin \frac{{2\pi }}{{15}}}} $

$= \frac{1}{{16}}$.

$ = \left( {1 + \cos \frac{\pi }{8} + \cos \frac{{7\pi }}{8} + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)$

$\left( {1 + \cos \frac{{5\pi }}{8} + \cos \frac{{3\pi }}{8} + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8} - \cos \frac{\pi }{8} + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)$

$\left( {1 + \cos \frac{{5\pi }}{8} - \cos \frac{{5\pi }}{8} + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \frac{1}{4}\,\,\left( {2 + 2\cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)\,\,\left( {2 + 2\cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$
$ = \frac{1}{4}\left( {2 + \cos \frac{{3\pi }}{4} + \cos \pi } \right)\left( {2 + \cos \frac{\pi }{4} + \cos \pi } \right)$

$ = \frac{1}{4}\,\left( {1 + \cos \frac{{3\pi }}{4}} \right)\,\left( {1 + \cos \frac{\pi }{4}} \right) = \frac{1}{4}\left( {1 - \cos \frac{\pi }{4}} \right)\,\left( {1 + \cos \frac{\pi }{4}} \right)$

$ = \frac{1}{4}\left( {1 - {{\cos }^2}\frac{\pi }{4}} \right) = \frac{1}{4}\left( {1 - \frac{1}{2}} \right) = \frac{1}{8}$.

वैकल्पिक​ : $\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 - \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 - \cos \frac{{3\pi }}{8}} \right)$

$ = \left( {1 - {{\cos }^2}\frac{\pi }{8}} \right){\rm{ }}\left( {1 - {{\cos }^2}\frac{{3\pi }}{8}} \right) = {\sin ^2}\frac{\pi }{8}{\sin ^2}\frac{{3\pi }}{8}$

$ = \frac{1}{4}{\left( {2\sin \frac{\pi }{8}.\sin \frac{{3\pi }}{8}} \right)^2}$$ = \frac{1}{4}{\left( {\cos \frac{\pi }{4} - \cos \frac{\pi }{2}} \right)^2} = \frac{1}{8}$.

$ = \frac{{2{{\sin }^2}(B/2)}}{{2\sin (B/2)\cos (B/2)}} = \tan \frac{B}{2}$

==> $\tan 2A = \tan B$.

$ = \frac{1}{2}\,\left[ {\frac{{\sqrt 5 + 1}}{4} - \frac{1}{2}} \right]\,\left[ {\frac{{\sqrt 5 + 1}}{4}} \right] $

$= \frac{1}{2}\,\left[ {\frac{{\sqrt 5 - 1}}{4}} \right]\,\left[ {\frac{{\sqrt 5 + 1}}{4}} \right]$

$ = \frac{{5 - 1}}{{32}} = \frac{4}{{32}} = \frac{1}{8}$.

$ {\sin ^2}\theta  + {\cos ^2}\theta {\cos ^2}\theta  \le {\sin ^2}\theta  + {\cos ^2}\theta $

(चूँकि ${\cos ^2}\theta  \le 1)$

$ \Rightarrow {\sin ^2}\theta  + {\cos ^4}\theta  \le 1 \Rightarrow A \le 1$

पुन: ${\sin ^2}\theta  + {\cos ^4}\theta  = 1 - {\cos ^2}\theta  + {\cos ^4}\theta $

$ = {\cos ^4}\theta  - {\cos ^2}\theta  + 1 = {\left( {{{\cos }^2}\theta  - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4}$

अत: $\frac{3}{4} \le A \le 1.$

अब ${\sin ^2}\alpha  + {\sin ^2}\beta  - {\sin ^2}\gamma $

$ = {\sin ^2}\alpha  + \sin (\beta  - \gamma )\sin (\beta  + \gamma )$

$ = {\sin ^2}\alpha  + \sin (\pi  - \alpha )\sin (\beta  + \gamma )$

                                                            $(\because \alpha  + \beta  - \gamma  = \pi )$ 

$ = {\sin ^2}\alpha  + \sin \alpha \sin (\beta  + \gamma ) = \sin \alpha \{ \sin \alpha  + \sin (\beta  + \gamma )\} $

$ = \sin \alpha \{ \sin (\pi  - \overline {\beta  + \gamma )}  + \sin (\beta  + \gamma )\} $

$ = \sin \alpha \{  - \sin (\gamma  - \beta ) + \sin (\gamma  + \beta )\} $

$ = \sin \alpha \{ 2\sin \beta \cos \gamma \}  = 2\sin \alpha \sin \beta \cos \gamma $.

$\Rightarrow \frac{\alpha }{2} + \frac{\beta }{2} + \frac{\gamma }{2} = \pi $

$ \Rightarrow \tan \left( {\frac{\alpha }{2} + \frac{\beta }{2} + \frac{\gamma }{2}} \right) = \tan \pi  = 0$

$ \Rightarrow \tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2} = 0$

$ \Rightarrow \tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} $

$= \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$.

.

एवं $\sin \,(\alpha  - \beta ) = \frac{5}{{13}}$

$ \Rightarrow \,\,\sin \,(\alpha  + \beta ) = \frac{3}{5}$

एवं $\cos \,(\alpha  - \beta ) = \frac{{12}}{{13}}$

$ \Rightarrow \,\,2\alpha  = {\sin ^{ - 1}}\frac{3}{5} + {\sin ^{ - 1}}\frac{5}{{13}}$

$ = {\sin ^{ - 1}}\left[ {\frac{3}{5}\sqrt {1 - \frac{{25}}{{169}}}  + \frac{5}{{13}}\sqrt {1 - \frac{9}{{25}}} } \right]$

$ \Rightarrow \,\,2\alpha  = {\sin ^{ - 1}}\,\left( {\frac{{56}}{{65}}} \right)\, $

$\Rightarrow \,\sin \,2\alpha  = \frac{{56}}{{65}}$

अब, $\tan \,2\alpha  = \frac{{\sin \,2\alpha }}{{\cos \,2\alpha }}$

$= \frac{{56/65}}{{33/65}} = \frac{{56}}{{33}}$.

$\left( \because   {\tan \theta  =  - \frac{4}{3}} \right)$

${\sin ^2}\theta  = \frac{1}{{{\rm{cose}}{{\rm{c}}^2}\theta }} = \frac{{16}}{{25}} \Rightarrow \sin \theta  =  \pm \frac{4}{5},$

दोनों मान हो सकते हैं, क्योंकि $\tan \theta  =  - \frac{4}{3}\,\,$

अर्थात् $\theta $ दूसरे व चौथे चतुर्थांश में है।

हम जानते हैं, $\tan \,(\alpha  + \beta ) = \frac{{\tan \,\alpha  + \tan \,\beta }}{{1 - \tan \,\alpha \,\tan \,\beta }}$

$ = \frac{{\frac{m}{{m + 1}} + \frac{1}{{2m + 1}}}}{{1 - \frac{m}{{(m + 1)}}\,\frac{1}{{(2m + 1)}}}} = \frac{{2{m^2} + m + m + 1}}{{2{m^2} + m + 2m + 1 - m}}$

$ = \frac{{2{m^2} + 2m + 1}}{{2{m^2} + 2m + 1}} = 1\,\, $

$\Rightarrow \,\,\tan \,(\alpha  + \beta ) = \tan \frac{\pi }{4}$

अत: $\alpha  + \beta  = \frac{\pi }{4}$.

ट्रिक : चूँकि $\alpha  + \beta $, $m$ से स्वतंत्र है।

अत: $m = 1$रखने पर, $\tan \,\alpha  = \frac{1}{2}$ व $\tan \,\beta  = \frac{1}{3}$, इसलिए

$\tan \,(\alpha  + \beta ) = \frac{{(1/2) + (1/3)}}{{1 - (1/6)}} = 1$ 

अत: $\alpha  + \beta  = \frac{\pi }{4}$

($m$ के किसी और मान के लिए भी जाँच करें )

$2{\sin ^2}\beta = 1 - \cos 2\beta $

$L.H.S.$ $ = - \cos 2\beta + 2\cos (\alpha + \beta )\,[2\sin \alpha \sin \beta + \cos (\alpha + \beta )]$ 

$ = - \cos 2\beta + 2\cos (\alpha + \beta )\cos (\alpha - \beta )$

$ = - \cos 2\beta + (\cos 2\alpha + \cos 2\beta ) = \cos 2\alpha $.

$ = \frac{{\sin \,\,{{20}^o}\sin \,\,{{40}^o}\sin \,\,{{80}^o}\tan {{60}^o}}}{{\cos \,\,{{20}^o}\cos \,\,{{40}^o}\cos \,\,{{80}^o}}}$

यहाँ  ${N^r} = (\sin \,\,{20^o}\sin \,\,{40^o}\sin \,\,{80^o})$

$ = \frac{{\sin \,\,{{20}^o}}}{2}\,(2\,\,\sin \,\,{40^o}\sin \,\,{80^o})$

$ = \frac{{\sin \,\,{{20}^o}}}{2}\,(\cos \,\,{40^o} - \cos \,\,{120^o})$

$ = \frac{1}{2}\sin \,\,{20^o}\,\left( {1 - 2\,\,{{\sin }^2}{{20}^o} + \frac{1}{2}} \right)$

$ = \frac{1}{2}\sin \,\,{20^o}\,\left( {\frac{3}{2} - 2\,\,{{\sin }^2}{{20}^o}} \right) = \frac{{\sin \,{{60}^o}}}{4} = \frac{{\sqrt 3 }}{8}$

अब, we take ${D^r} = \cos {20^o}\cos {40^o}\cos {80^o}$

$ = \frac{{\sin \,\,{2^3}\,{{20}^o}}}{{{2^3}\,\sin \,\,{{20}^o}}} = \frac{{\sin \,\,{{160}^o}}}{{8\,\,\sin \,\,{{20}^o}}} = \frac{{\sin \,\,{{20}^o}}}{{8\,\,\sin \,\,{{20}^o}}} = \frac{1}{8}$

$\therefore $ अतः $\tan \,\,{20^o}\tan \,\,{40^o}\tan \,\,{80^o} = \frac{{\sqrt 3 /8}}{{1/8}}$

इसलिए $\tan {20^o}\tan {40^o}\tan {60^o}\tan {80^o} = \sqrt 3 .\sqrt 3 = 3$.

$ = \frac{{2\left( {\frac{{\cos 10^\circ }}{2} - \frac{{\sqrt 3 }}{2}\sin 10^\circ } \right)}}{{(2\sin 10^\circ \cos 10^\circ ) \times \frac{1}{2}}}$

$ = \frac{{4\,\sin \,({{30}^o} - {{10}^o})}}{{\sin \,{{20}^o}}} $

$= \frac{{4\,\sin \,{{20}^o}}}{{\sin \,{{20}^o}}} = 4$.

$\therefore \,\,\,{m^2} - {n^2} = 4\,\tan \theta \,.\,\sin \theta $…..$(i)$

$4\sqrt {mn}  = 4\sqrt {{{\tan }^2}\theta  - {{\sin }^2}\theta }  $

$= 4\,\sin \theta \,.\,\tan \theta $…..$(ii)$

$(i)$ व $(ii)$ से, ${m^2} - {n^2} = 4\sqrt {mn} $.

अतः $A + C = 180^\circ \Rightarrow A = 180^\circ - C$

$ \Rightarrow \cos A = \cos (180^\circ - C) = - \cos C$

$ \Rightarrow \cos A + \cos C = 0$.....$(i)$

इसी प्रकार, $\cos B + \cos D = 0$ .....$(ii)$

जोड़ने पर, $\cos A + \cos B + \cos C + \cos D = 0.$

अब, $\tan \,(A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A\,\tan B}} $

$= \frac{{ - \frac{1}{2} - \frac{1}{3}}}{{1 - \frac{1}{2}.\frac{1}{3}}} =  - 1$

$ \Rightarrow \,\,\tan \,(A + B) = \tan \frac{{3\pi }}{4}$

अत: $A + B = \frac{{3\pi }}{4}.$

${\sec ^2}\theta  = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $

$\Rightarrow 4xy \ge {(x + y)^2} $

$\Rightarrow {(x - y)^2} \le 0$

$x = y$,  $(x,\,y \in R)$

$= \frac{{2{{\cos }^2}A}}{{2\sin A\cos A}} = \frac{{1 + \cos 2A}}{{\sin 2A}}$

$A = 7\frac{{{1^o}}}{2}$ रखने पर

$ \Rightarrow \cot 7\frac{{{1^o}}}{2} = \frac{{1 + \cos {{15}^o}}}{{\sin {{15}^o}}}$

सरल करने पर,

$\cot 7\frac{{{1^o}}}{2} = \sqrt 6  + \sqrt 2  + \sqrt 3  + \sqrt 4 $.

$ \Rightarrow \,\,\frac{{m + n}}{{m - n}} $

$= \frac{{\tan \,(\theta + {{120}^o}) + \tan \,(\theta - {{30}^o})}}{{\tan \,(\theta + {{120}^o}) - \tan \,(\theta - {{30}^o})}}$

                                                             (योगान्तरानुपात से)

$ = \frac{{\sin (\theta + {{120}^o})\cos (\theta - {{30}^o}) + \cos (\theta + {{120}^o})\sin (\theta - {{30}^o})}}{{\sin (\theta + {{120}^o})\cos (\theta - {{30}^o}) - \cos (\theta + {{120}^o})\sin (\theta - {{30}^o})}}$

$ = \frac{{\sin \,(2\theta + {{90}^o})}}{{\sin \,({{150}^o})}} $

$= \frac{{\cos \,2\theta }}{{1/2}} = 2\,\cos \,2\theta $.

$ = \frac{{\sin \,(B + A) + \sin \,({{90}^o} - \overline {B - A} )}}{{\sin \,(B - A) + \sin \,({{90}^o} - \overline {A + B} )}}$

$ = \,\frac{{2\,\sin \,(A + {{45}^o})\,\cos \,({{45}^o} - B)}}{{2\,\sin \,({{45}^o} - A)\,\cos \,({{45}^o} - B)}}$

$ = \frac{{\sin \,(A + {{45}^o})}}{{\sin \,({{45}^o} - A)}} $

$= \frac{{\cos A + \sin A}}{{\cos A - \sin A}}$.

$ = {\sin ^2}{36^o}\,\,{\sin ^2}\,{72^o} = \frac{1}{4}\,\left\{ {(2\,\,{{\sin }^2}{{36}^o})\,\,(2\,\,{{\sin }^2}\,\,{{72}^o})} \right\}$

$ = \frac{1}{4}\left\{ {(1 - \cos \,\,{{72}^o})\,\,(1 - \cos \,\,{{144}^o})} \right\}$

$ = \frac{1}{4}\left\{ {(1 - \sin \,\,{{18}^o})\,\,(1 + \cos \,\,{{36}^o})} \right\}$

$ = \frac{1}{4}\left[ {\left( {1 - \frac{{\sqrt 5 - 1}}{4}} \right)\,\,\left( {1 + \frac{{\sqrt 5 + 1}}{4}} \right)} \right]$

$= \frac{{20}}{{16}} \times \frac{1}{4} = \frac{5}{{16}}$.

$ = 2{\cos ^2}28^\circ + 2\sin 62^\circ .\sin 4^\circ $

$ = 2{\cos ^2}28^\circ + 2\cos 28^\circ .\sin 4^\circ $

$ = 2\cos 28^\circ (\cos 28^\circ + \cos 86^\circ )$ 

$ = 2\cos 28^\circ .2\cos 57^\circ \cos 29^\circ $

$ = 4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.

वैकल्पिक : प्रतिबन्धित सर्वसमिका के उपयोग से,

$\cos A + \cos B - \cos C = - 1 + 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$                                                              $[\, \because 56^\circ  + 58^\circ  + 66^\circ  = 180^\circ ]$

अभीष्ट व्यंजक का मान  $4\cos 28^\circ \cos 29^\circ \sin 33^\circ $ होगा। 

$ = {e^{{{\log }_{10}}\,(\tan \,\,{1^o}\,\tan \,\,{2^o}\,\,\tan \,\,{3^o}.....\tan \,\,{{89}^o})}} = {e^{{{\log }_{10}}\,\,1}} = {e^o}$

$= 1$

यहाँ ${x^2} + \frac{1}{{{x^2}}} \ge 2$

अत: $f(x) = {\cos ^2}x + \frac{1}{{{{\cos }^2}x}} \ge 2$

लेकिन $ - 1 \le \sin \left( {\theta  + \frac{\pi }{2}} \right) \le 1$

$\Rightarrow  - \sqrt 2  \le \sqrt 2 \sin \left( {\theta  + \frac{\pi }{4}} \right) \le \sqrt 2 $

अत: $(\sin \theta  + \cos \theta )$ का अधिकतम मान

अर्थात्, $\sqrt 2 \sin \left( {\theta  + \frac{\pi }{4}} \right) = \sqrt 2 $ है।

$\therefore $$\sin \left( {\theta  + \frac{\pi }{4}} \right) = 1 $

$\Rightarrow \sin \left( {\theta  + \frac{\pi }{4}} \right) = \sin \frac{\pi }{2}$

$\theta  + \frac{\pi }{4} = \frac{\pi }{2} $

$\Rightarrow \theta  = \frac{\pi }{4} = {45^o}$.

अत: फलन $\sqrt 3 \sin x + \cos x$ के ग्राफ की $x-$ अक्ष से अधिकतम दूरी $2$ है।

$ \Rightarrow A = 180^\circ  - C$

$ \Rightarrow \cos A = \cos (180^\circ  - C) =  - \cos C$

$ \Rightarrow \cos A + \cos C = 0$.....$(i)$

अब $B + D = 180^\circ ,$ तब $\cos B + \cos D = 0$.....$(ii)$

समीकरण $(i)$ में से $(ii)$ को घटाने पर,

$\cos A - \cos B + \cos C - \cos D = 0$.

वैकल्पिक : $2\,\left( {\frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x} \right) = 2\sin \,\left( {x + \frac{\pi }{6}} \right)$

चूँकि $\sin x$, $x = \frac{\pi }{2}$ पर महत्तम है

अत: $x + \frac{\pi }{6} = \frac{\pi }{2}$ या  $x = \frac{\pi }{3}$.

$ \Rightarrow \frac{{9{{\tan }^2}\theta + 4{{\cot }^2}\theta }}{2} \ge \sqrt {4{{\cot }^2}\theta .9{{\tan }^2}\theta } $

$ \Rightarrow 9{\tan ^2}\theta + 4{\cot ^2}\theta \ge 12$

अतः न्यूनतम मान $12$ है।

$\therefore $ ${\sin ^2}x + 3$ का अधिकतम मान 4 है।

.

चूँकि $ - 1 \le \cos \left( {\theta  - \frac{\pi }{4}} \right) \le 1$

$\Rightarrow$ $ - \sqrt 2  \le \sqrt 2 \cos \left( {\theta  - \frac{\pi }{4}} \right) \le \sqrt 2 $

अत: $f(x)$ का न्यूनतम मान  $ - \sqrt 2 $ है।

हम जानते हैं कि, $ - 1 \le \sin 2x \le 1 \Rightarrow  - \frac{1}{2} \le \frac{1}{2}\sin 2x \le \frac{1}{2}$

अत: $f(x)$ के उच्चिष्ठ व निम्निष्ठ मान क्रमश: $\frac{1}{2}$ व $ - \frac{1}{2}$ हैं।

अत: $3\cos x + 4\sin x + 5$ का न्यूनतम मान

$ =  - 5 + 5 = 0$.

$ \Rightarrow f(x) = 2\left( {\frac{{\sqrt 3 }}{2}\cos x + \frac{1}{2}\sin x} \right) = 2\sin \left( {x + \frac{\pi }{3}} \right)$

लेकिन$ - 1 \le \sin \left( {x + \frac{\pi }{3}} \right) \le 1$

अत: $f(x)$ अधिकतम है यदि

$x + \frac{\pi }{3} = 90^\circ  \Rightarrow x = 30^\circ $.

$\Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 \ge 0$

$x = \tan \theta $ रखने पर, ${\tan ^2}\theta  + {\cot ^2}\theta  \ge 2$.

$ = \left\{ {\cos \left( {\frac{\pi }{3} - x} \right) + \cos \left( {\frac{\pi }{3} + x} \right)} \right\}\left\{ {\cos \left( {\frac{\pi }{3} - x} \right) - \cos \left( {\frac{\pi }{3} + x} \right)} \right\}$

$ = \left\{ {2\cos \frac{\pi }{3}\cos x} \right\}\left\{ {2\sin \frac{\pi }{3}\sin x} \right\}$

$ = \sin \frac{{2\pi }}{3}\sin 2x = \frac{{\sqrt 3 }}{2}\sin 2x$

अधिकतम मान $ = \frac{{\sqrt 3 }}{2}$ है, $\{  - 1 \le \sin 2x \le 1\} $

$\therefore f(\theta ) \ge 4 + 0$         $( \because {\sin ^2}\theta  \ge 0)$ 

$\therefore $$f(\theta )$ का न्यूनतम मान $4$ है

$f(\theta ) = r.(\cos \alpha \cos \theta  + \sin \alpha \sin \theta ) = 5.\cos (\theta  - \alpha )$

$\therefore $ $f(\theta )$ का उच्चिष्ठ मान $ = 5.1 = 5$

{चूँकि $\cos (\theta  - \alpha )$का उच्चिष्ठ मान $1$ है}

वैकल्पिक : चूँकि हम जानते हैं कि $a\sin \theta  + b\cos \theta $ का अधिकतम मान $ + \sqrt {{a^2} + {b^2}} $

तथा न्यूनतम मान $ - \sqrt {{a^2} + {b^2}} $है।

अत: $(3\cos \theta  + 4\sin \theta )$ का अधिकतम मान $ + \sqrt {{3^2} + {{( - 4)}^2}}  = 5$

एवं न्यूनतम मान $-5$ है।

$ \Rightarrow $$\tan 2A = \frac{{\tan (A + B) + \tan (A - B)}}{{1 - \tan (A + B)\tan (A - B)}}$

$= \frac{{p + q}}{{1 - pq}}$.

$ = \sin \alpha + \sin \beta + \sin \gamma - \sin \alpha \cos \beta \cos \gamma $ 

$ - \cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \sin \alpha \sin \beta \sin \gamma $

$ = \sin \alpha (1 - \cos \beta \cos \gamma ) + \sin \beta (1 - \cos \alpha \cos \gamma )$

$ + \sin \gamma (1 - \cos \alpha \cos \beta ) + \sin \alpha \sin \beta \sin \gamma > 0$ 

$\therefore \sin \alpha + \sin \beta + \sin \gamma > \sin (\alpha + \beta + \gamma )$ 

$ \Rightarrow \frac{{\sin (\alpha + \beta + \gamma )}}{{\sin \alpha + \sin \beta + \sin \gamma }} < 1$ .

==> $x = \cos {40^o}\cos {80^o},\,\,\,y = - \sin {80^o}\sin {40^o}$

अतः  $x > 0$ एवं $y < 0$ एवं $xy < 0$

अब $z = 1 + xy$ ==> $0 < z < 1$.

अब  ${x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3x\frac{1}{x}\left( {x + \frac{1}{x}} \right)$ 

$=  {(2\cos \theta )^3} - 3(2\cos \theta ) = 8{\cos ^3}\theta - 6\cos \theta $

$=  2(4{\cos ^3}\theta - 3\cos \theta ) = 2\cos 3\theta $. 

ट्रिक​ : $x = 1$ $ \Rightarrow $ $\theta = {0^\circ }$. 

तब  ${x^3} + \frac{1}{{{x^3}}} = 2 = 2\cos 3\theta $.

$\cos 2A + \cos 2B + \cos 2C + 4\sin A\,\sin B\,\sin C$

$ = \cos {180^o} + \cos {180^o} + \cos {180^o} + 4\sin {90^o}\sin {90^o}\sin {90^o}$

$ = - 1 - 1 - 1 + 4 \cdot 1\cdot 1\cdot 1 = - 3 + 4 = 1$.

$ = 2\sin A\cos A + 2\cos (B + C)\sin (B - C)$

$\{ \because A + B + C = \pi ,\,\therefore \,B + C = \pi  - A,\cos (B + C) = \cos (\pi  - A),$ $\cos (B + C) = - \cos A,\,\sin (B + C) = \sin A\} $

$\cos (B + C) =  - \cos A,\,\sin (B + C) = \sin A\} $

$ = 2\cos A\,\,[\sin A - \sin (B - C)]$

$ = 2\cos A\,[\sin (B + C) - \sin (B - C)]$

$ = 2\cos A.2\cos B.\sin C$

$ = 4\cos A.\,\cos B.\,\sin C$.

ट्रिक: $A = B = C = {60^o}$ रखने पर इन मानों के लिए विकल्प $(a)$ तथा $(d)$ सन्तुष्ट होते हैं,

अब केवल $A = B = 45^\circ $ तथा $c = {90^o}$ रखने पर केवल विकल्प $(d)$ सन्तुष्ट होता है।

अत: विकल्प $(d)$ सही उत्तर है।

$ = 2\cos (x + y)\cos (x - y) - 2{\cos ^2}z + 1$ 

$ = 2\cos (\pi - z)\cos (x - y) - 2{\cos ^2}z + 1$ 

$ = 1 - 2\cos z\{ \cos (x - y) - \cos (x + y)\} $

$ = 1 - 2\cos z2\sin x\sin y = 1 - 4\sin x\sin y\cos z$.

अब, $\sin 2A + \sin 2B + \sin 2C$

$ = 2\sin (A + B)\cos (A - B) + 2\sin C\cos C$

$ = 2\sin (\pi  - C)\cos (A - B) + 2\sin C\cos (\pi  - \overline {A + B} )$

$ = 2\sin C\cos (A - B) - 2\sin C\cos (A + B)$

$ = 2\sin C\{ \cos (A - B) - \cos (A + B)\} $

$ = 2\sin C\{ 2\sin A\sin B\}  = 4\sin A\sin B\sin C$.

$ \Rightarrow \sin A + \sin B + \sin C $

$= 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}$ 

$ = 2\sin \left( {\frac{\pi }{2} - \frac{C}{2}} \right)\cos \frac{{A - B}}{2} + 2\cos \frac{C}{2}\sin \left( {\frac{\pi }{2} - \frac{{\overline {A + B} }}{2}} \right)$

$ = 2\cos \frac{C}{2}\cos \frac{{A - B}}{2} + 2\cos \frac{C}{2}\cos \frac{{A + B}}{2}$ 

$ = 2\cos \frac{C}{2}\left[ {\cos \frac{{A - B}}{2} + \cos \frac{{A + B}}{2}} \right]$

$ = 2\cos \frac{C}{2}\left( {2\cos \frac{A}{2}\cos \frac{B}{2}} \right) $

$= 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$ .

$ = \sqrt {\frac{{2{{\sin }^2}\left( {\frac{\pi }{4} - \frac{A}{2}} \right)}}{{2{{\cos }^2}\left( {\frac{\pi }{4} - \frac{A}{2}} \right)}}} = \tan \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$.

$ = {(3\cos A - 4{\cos ^3}A)^2} + {(3\sin A - 4{\sin ^3}A)^2}$

$ = {(\cos 3A)^2} + {(\sin 3A)^2} = 1$.

ट्रिक : $A = \frac{\pi }{2},{0^o}$ रखने पर व्यंजक का मान $1$ आता है।

अत: यह  $A$ से स्वतंत्र व $1$ के बराबर है।

==> ${\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \frac{1}{{25}}$ 

$\sin 2x = \frac{{ - 24}}{{25}}$

==> $\cos 2x = \frac{{ - 7}}{{25}}$ 

==> $\tan 2x = \frac{{24}}{7}$.

$\Rightarrow 1 + {\tan ^2}\theta  = 2\,(1 + {\tan ^2}\phi )$

$\Rightarrow$ ${\sec ^2}\theta  = 2{\sec ^2}\phi $

$\Rightarrow {\cos ^2}\phi = 2{\cos ^2}\theta $

$\Rightarrow$ ${\cos ^2}\phi = 1 + \cos 2\theta  $

$\Rightarrow {\sin ^2}\phi   + \cos 2\theta  = 0$.

ट्रिक : माना $\theta  = {45^o}$ तब $\phi = 0$

$\therefore \;\cos (2 \times {45^o}) + {\sin ^2}0 = 0 + 0 = 0$.

==> $2\cos 2\theta = 1$ 

==> $\cos 2\theta = \frac{1}{2} = \cos {60^o}$

==> $2\theta = {60^o}$

$\Rightarrow \theta = {30^o}$.

दोनों तरफ वर्ग करने पर,

$\Rightarrow 1 + \sin 2A = 2\, \Rightarrow \sin 2A = 1 = \sin {90^o}$

$\Rightarrow 2A = {90^o}$या  $A = {45^o}$

अब ${\cos ^2}A = {(\cos {45^o})^2}$

$= {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}$.

$ \Rightarrow \frac{{1 - {{\tan }^2}\frac{\theta }{2}}}{{1 + {{\tan }^2}\frac{\theta }{2}}} = \cos \theta $.

$\Rightarrow$ $\tan A =  - \frac{4}{3}$, $({90^o} < A < {180^o})$

$\tan A = \frac{{2\tan \frac{A}{2}}}{{1 - {{\tan }^2}\frac{A}{2}}}$, (माना $\tan \frac{A}{2} = P$)

$ - \frac{4}{3} = \frac{{2P}}{{1 - {P^2}}}$

$\Rightarrow$  $4{P^2} - 6P - 4 = 0$

$\Rightarrow$ $P = \frac{{ - 1}}{2}\,$(असम्भव),

अत:  $\tan \frac{A}{2} = 2$.

$\therefore \cos 3\theta = 4\frac{1}{{{2^3}}}{\left( {a + \frac{1}{a}} \right)^3} - 3\frac{1}{2}\left( {a + \frac{1}{a}} \right)$

$ \Rightarrow \cos 3\,\theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right)\,\left[ {{{\left( {a + \frac{1}{a}} \right)}^2} - 3} \right]$ 

==> $\cos 3\theta = \frac{1}{2}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$.

$\tan 2\theta + \sec 2\theta = \frac{{2t}}{{1 - {t^2}}} + \frac{{1 + {t^2}}}{{1 - {t^2}}} $

$= \frac{{{{(1 + t)}^2}}}{{(1 - t)(1 + t)}} = \frac{{1 + t}}{{1 - t}}$.

$\cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha } $     [$\because \alpha$ तृतीय चतुर्थांश में है]

$ =  - \sqrt {1 - \frac{9}{{25}}}  =  - \frac{4}{5}$

$\therefore \,\,\,\cos (\alpha /2) =  - \sqrt {\frac{{1 - \frac{4}{5}}}{2}}  =  - \frac{1}{{\sqrt {10} }}$.

$= \frac{{\sin (A/2)}}{{\cos (A/2)}} $

$= \pm \sqrt {\frac{{(1 - \cos A)/2}}{{(1 + \cos A)/2}}} $

$= \pm \sqrt {\frac{{1 - \cos A}}{{1 + \cos A}}} $.

$ = \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}^2}}}{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)\,\left( {\cos \frac{A}{2} - \sin \frac{A}{2}} \right)}} $

$= \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}}$

$ = \frac{{1 + \tan \frac{A}{2}}}{{1 - \tan \frac{A}{2}}}$, (अंश तथा हर को ${\cos \frac{A}{2}}$ से भाग देने पर )

$ = \tan \left( {\frac{\pi }{4} + \frac{A}{2}} \right)$.

$L.H.S.$ $ = 16(\sin 3A - \sin 2A)$ 

$ = 16\sin A(3 - 4{\sin ^2}A - 2\cos A)$ 

$ = 16.\frac{{\sqrt 7 }}{4}\left( {3 - 4.\frac{7}{{16}} - 2.\frac{3}{4}} \right) = - \sqrt 7 $.

$= \frac{{\tan \alpha - \tan \beta }}{{\tan \alpha + \tan \beta }} $

$= \frac{{\sin (\alpha - \beta )}}{{\sin (\alpha + \beta )}}$.

$ = \frac{{1 - \cos 8A}}{{\cos 8A}}.\frac{{\cos 4A}}{{1 - \cos 4A}}$

$ = \frac{{2{{\sin }^2}4A}}{{\cos 8A}}\frac{{\cos 4A}}{{2{{\sin }^2}2A}}$

$ = \frac{{2\sin 4A\cos 4A}}{{\cos 8A}}\frac{{\sin 4A}}{{2{{\sin }^2}2A}}$

$ = \tan 8A\frac{{2\sin 2A\cos 2A}}{{2{{\sin }^2}2A}} $

$= \frac{{\tan 8A}}{{\tan 2A}}.$

$ = \frac{{\sin \theta + 2\sin \theta \cos \theta }}{{2{{\cos }^2}\theta + \cos \theta }} $

$= \frac{{\sin \theta (1 + 2\cos \theta )}}{{\cos \theta (1 + 2\cos \theta )}} $

$= \tan \theta $.

ट्रिक : $\theta  = 30^\circ $ रखने पर, चूंकि $\theta  = 30^\circ $ के लिए कोई भी विकल्प उभयनिष्ठ मान नहीं देता है।

$ = 2\sin A\cos A({\cos ^2}A - {\sin ^2}A)$ 

$ = 2\sin A\cos A\cos 2A $

$= \sin 2A\cos 2A $

$= \frac{1}{2}\sin 4A$.

$= \left( {\frac{{1 + {{\tan }^2}A}}{{1 - {{\tan }^2}A}} + 1} \right)\,(1 + {\tan ^2}A)$

$ = \frac{{2\,(1 + {{\tan }^2}A)}}{{1 - {{\tan }^2}A}}$

$= 2\sec 2A.$

$= \frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2}}}$

$ = \tan \frac{x}{2}$.

$ \Rightarrow \tan 3A = \frac{{3\tan A - {{\tan }^3}A}}{{1 - {{3+an }^2}A}} $

$= \frac{{3.\frac{1}{2} - \frac{1}{8}}}{{1 - 3.\frac{1}{4}}} $

$= \frac{{12 - 1}}{2} = \frac{{11}}{2}$.

$ = \frac{{\sin 3A - \sin A}}{{\cos A - \cos 3A}}$

$=\frac{{2\cos 2A\sin A}}{{2\sin 2A\sin A}}$

$= \frac{{\cos 2A}}{{\sin 2A}} = \cot 2A$.

$= \cos \left( {\frac{\pi }{2} + 2\theta } \right) $

$= - \sin 2\theta $.

अब $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} $

$= \sqrt {\frac{{1 + b/a}}{{1 - b/a}}} + \sqrt {\frac{{1 - b/a}}{{1 + b/a}}} $

$ = \frac{2}{{\sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} }} = \frac{2}{{\sqrt {1 - {{\tan }^2}x} }} $

$= \frac{2}{{\sqrt {1 - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} }} $

$= \frac{{2\cos x}}{{\sqrt {\cos 2x} }}$.

$ = {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta + {\sin ^2}\alpha + $

${\sin ^2}\beta + 2\sin \alpha \sin \beta $

$ = 2\{ 1 + \cos (\alpha - \beta )\}$

$= 4{\cos ^2}\left( {\frac{{\alpha - \beta }}{2}} \right)$.

$= \alpha \cos \theta + \beta {\cos ^3}\theta $

लेकिन  $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta $

==> $(\alpha ,\beta ) = ( - 3,\,4)$.

$=\sqrt {2 + \sqrt {2.2{{\cos }^2}2\theta } } $

$ = \sqrt {2 + 2\cos 2\theta }$

$= \sqrt {4{{\cos }^2}\theta } $

$= 2\cos \theta $.

$= \frac{1}{{\sin A}} - \frac{{2\cos A\cos 2A}}{{\sin 2A}}$

$ = \frac{1}{{\sin A}} - \frac{{2\cos A\cos 2A}}{{2\sin A\cos A}} $

$= \frac{{1 - \cos 2A}}{{\sin A}} $

$= \frac{{2{{\sin }^2}A}}{{\sin A}}$

$ = 2\sin A$.

$=\frac{1}{{\tan 3A - \tan A}} + \frac{{\tan A\tan 3A}}{{\tan 3A - \tan A}}$

$= \frac{1}{{\tan 2A}} = \cot 2A$.

$ = \frac{{2\sin A\cos A}}{{2{{\cos }^2}A}}\frac{{\cos A}}{{1 + \cos A}}$

$= \frac{{\sin A}}{{1 + \cos A}} $

$= \tan \frac{A}{2}$.

अब $a\cos 2\theta  + b\sin 2\theta  $

$= a\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) + b\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$\tan \theta  = \frac{b}{a}$ रखने पर,

$ = a\left( {\frac{{1 - \frac{{{b^2}}}{{{a^2}}}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} \right) + b\left( {\frac{{2\frac{b}{a}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} \right) $

$= a\left( {\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}} \right) + b\left( {\frac{{2ba}}{{{a^2} + {b^2}}}} \right)$

$ = \frac{1}{{({a^2} + {b^2})}}\{ {a^3} - a{b^2} + 2a{b^2}\} $

$= \frac{{a({a^2} + {b^2})}}{{{a^2} + {b^2}}} = a$.

$ = 2.2\sin \theta \cos \theta (1 - 2{\sin ^2}\theta )$

$ = 4\sin \theta (1 - 2{\sin ^2}\theta )\sqrt {1 - {{\sin }^2}\theta } $

$= \left[ {\frac{{\sin \left( {{2^3}.\frac{\pi }{7}} \right)}}{{{2^3}\sin \left( {\frac{\pi }{7}} \right)}}} \right] $

$= \frac{{\sin \frac{{8\pi }}{7}}}{{8\sin \frac{\pi }{7}}}$

$= - \frac{1}{8}$.

$ = (1 + \cos \,6x) + (\cos \,2x + \cos \,4x)$

$ = 2\,{\cos ^2}3x + 2\,\cos \,3x\,\cos x $

$= 2\,\cos \,3x\,(\cos \,3x + \cos \,x)$

$ = 4\,\cos x\,\cos \,2x\,\cos \,3x$.

$ = 2\cos x\,2{\sin ^2}2x$$ = 4\,\cos x\,{\sin ^2}\,2x $

$= 16\,{\sin ^2}x\,{\cos ^3}x$.

$\alpha = \beta = \gamma = {60^o} $ रखने पर

$\Rightarrow \frac{1}{2}(0) + \frac{1}{2}(0) + \frac{1}{2}(0) = 0$.

$ = \cos A + 2\cos {240^o}\cos A$

$ = \cos A\{ 1 + 2\cos ({180^o} + {60^o})\} $

$= \cos A\,\left\{ {1 + 2\,\left( { - \frac{1}{2}} \right)} \right\}$

$ = 0$.

$ = - \sin \,{60^o}\,\cos \,{30^o} - \sin \,{30^o}\,\cos \,{60^o}$

$ = - \left\{ {\sin \,({{60}^o} + {{30}^o})} \right\} = - 1$.

$ = \sin ({180^o} - {17^o})\cos ({360^o} - {13^o}) + \cos ({90^o} - {17^o})\sin ({180^o} - {13^o})$

$ = \sin {17^o}\cos {13^o} + \cos {17^o}\sin {13^o} $

$= \sin {30^o} = 1/2$.

$ = \frac{{(\sin \,3\theta + \sin \,9\theta ) + (\sin \,5\theta + \sin \,7\theta )}}{{(\cos \,3\theta + \cos \,9\theta ) + (\cos \,5\theta + \cos \,7\theta )}}$

$ = \frac{{2\,\sin \,6\theta \,\cos \,3\theta + 2\,\sin \,6\theta \,\cos \,\theta }}{{2\,\cos \,6\theta \,\cos \,3\theta + 2\,\cos \,6\theta \,\cos \,\theta }}$

$ = \frac{{2\,\sin \,6\theta \,(\cos \,3\theta + \cos \theta )}}{{2\,\cos \,6\theta \,(\cos \,3\theta + \cos \theta )}}$

$= \tan \,6\theta $.

$ \Rightarrow \,\,\tan \,\,3A - \tan \,\,2A - \tan A = \tan \,\,3A\,\tan \,\,2A\,\,\tan A$.

$ = \frac{{\sin {{160}^o}}}{{8\sin {{20}^o}}} = \frac{1}{8}$.

$ = \frac{{1 - \tan \,\,{{12}^o}}}{{1 + \tan \,\,{{12}^o}}} + \tan \,\,{147^o}$

$ = \tan \,\,{33^o} - \tan \,\,{33^o} = 0$.

$ = \frac{{\sin \frac{{{2^4}\pi }}{5}}}{{{2^4}\sin \frac{\pi }{5}}} = \frac{{\sin \frac{{16\pi }}{5}}}{{16\,\sin \frac{\pi }{5}}} $

$= \frac{{\sin \,\left( {3\pi + \frac{\pi }{5}} \right)}}{{16\,\sin \frac{\pi }{5}}}$

$ = \frac{{ - \sin \frac{\pi }{5}}}{{16\,\sin \frac{\pi }{5}}} = - \frac{1}{{16}}$.

$\Rightarrow \tan 5x = \tan (3x + 2x)$

==> $\tan 5x = \frac{{\tan 3x + \tan 2x}}{{1 - \tan 3x\tan 2x}}$

==>$\tan 5x - \tan 5x\tan 3x\tan 2x = \tan 3x + \tan 2x$

==> $\tan 5x\tan 3x\tan 2x = \tan 5x - \tan 3x - \tan 2x$.

$= \sqrt 2 \,.\,\cos \,\,{60^o}$

$ = \sqrt 2 \,.\frac{1}{2} = \frac{1}{{\sqrt 2 }}$.

$ = (2 + \sqrt 3 ) - (2 - \sqrt 3 ) = 2\sqrt 3 $.

इसलिए $2A = 90^\circ - 2B$

$\therefore \cos 2A = \sin 2B$.

$ = \frac{{1 - 1/\sqrt 3 }}{{1 + 1/\sqrt 3 }} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} $

$= 2 - \sqrt 3 $.

$\frac{{1 + \cos A}}{{1 - \cos A}}$

$ = \frac{{2{{\cos }^2}\frac{A}{2}}}{{2{{\sin }^2}\frac{A}{2}}} $

$= {\cot ^2}\frac{A}{2} = {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}$.

$\sec \theta  = \frac{{1 + {{\tan }^2}(\theta /2)}}{{1 - {{\tan }^2}(\theta /2)}}$

$\Rightarrow \frac{5}{4} = \frac{{1 + {{\tan }^2}(\theta /2)}}{{1 - {{\tan }^2}(\theta /2)}}$

$\Rightarrow 5 - 5{\tan ^2}(\theta /2) = 4 + 4{\tan ^2}(\theta /2)$

$\Rightarrow 9{\tan ^2}(\theta /2) = 1\, $

$\Rightarrow \tan (\theta /2) = \frac{1}{3}$.

$= \frac{{\frac{{{{\cos }^2}{{15}^o}}}{{{{\sin }^2}{{15}^o}}} - 1}}{{\frac{{{{\cos }^2}{{15}^o}}}{{{{\sin }^2}{{15}^o}}} + 1}}$

$ = \frac{{{{\cos }^2}{{15}^o} - {{\sin }^2}{{15}^o}}}{{{{\cos }^2}{{15}^o} + {{\sin }^2}{{15}^o}}}$

$= \cos ({30^o}) = \frac{{\sqrt 3 }}{2}$.

$L.H.S.$ $ = 2\,\sin \gamma \cos \,(\beta - \alpha ) + 2\,\sin \,( - \gamma )\,\cos \,(\alpha + \beta )$

$ = 2\,\sin \,\gamma \,[\cos \,(\beta - \alpha ) - \cos \,(\alpha + \beta )]$

$ = 2\,\sin \,\gamma \,.\,2\,\sin \alpha \,\sin \beta $

$ = 4\sin \alpha \sin \beta \sin \gamma $.

$ \Rightarrow \,\,\frac{{\sin \,(x + y) + \sin \,(x - y)}}{{\sin \,(x + y) - \sin \,(x - y)}} $

$= \frac{{(a + b) + (a - b)}}{{(a + b) - (a - b)}}$

$ \Rightarrow \,\,\frac{{2\,\sin x\,\cos y}}{{2\,\cos x\,\sin y}} = \frac{{2a}}{{2b}}\, $

$\Rightarrow \,\,\frac{{\tan x}}{{\tan y}} = \frac{a}{b}$.

$\Rightarrow \,\frac{a}{b} = \frac{{\sin \,\alpha }}{{\sin \,(\alpha  + 2\beta )}}$

$ \Rightarrow \,\,\frac{{a + b}}{{a - b}} $

$= \frac{{\sin \,\alpha  + \sin \,(\alpha  + 2\beta )}}{{\sin \,\alpha  - \sin \,(\alpha  + 2\beta )}} $

$= \frac{{2\,\sin \,(\alpha  + \beta )\,\cos \,\beta }}{{ - 2\,\cos \,(\alpha  + \beta )\,\sin \,\beta }}$

$ =  - \tan \,(\alpha  + \beta )\,\cot \,\beta $

$=  - \frac{{\cot \beta }}{{\cot \,(\alpha  + \beta )}}$.

$ = \cos \left( {\frac{\pi }{6} + \theta + \frac{\pi }{6} - \theta } \right)\cos \left( {\frac{\pi }{6} + \theta - \frac{\pi }{6} + \theta } \right)$

                                                 $[ \because {\cos ^2}A - {\sin ^2}B = \cos (A + B)\cos (A - B)]$ 

$ = \cos \frac{{2\pi }}{6}\cos 2\theta = \frac{1}{2}\cos 2\theta $.

$ = \cos \,\left( {\frac{\pi }{4} - \beta + \alpha - \frac{\pi }{4}} \right)\,\cos \,\left( {\frac{\pi }{4} - \beta - \alpha + \frac{\pi }{4}} \right)\,$

$ = \cos (\alpha - \beta )\cos \left( {\frac{\pi }{2} - \overline {\alpha + \beta } } \right) $

$= \cos (\alpha - \beta )\sin (\alpha + \beta )$.

$= \frac{{1 + \tan A\,\,\tan B}}{{\tan A - \tan B}}$

$ = \frac{1}{{\tan A - \tan B}} + \frac{{\tan A\,\,\tan B}}{{\tan A - \tan B}} $

$= \frac{1}{x} + \frac{1}{y}$.

$\cos \theta  = \sqrt {1 - {{\sin }^2}\theta }  = \sqrt {1 - {{\left( {\frac{{12}}{{13}}} \right)}^2}}  = \frac{5}{{13}}$

एवं $\cos \phi = \frac{{ - 3}}{5},\sin \phi = \sqrt {1 - \frac{9}{{25}}}  = \frac{{ - 4}}{5}$,

                              $\left[ \because {\pi  < \phi < \frac{{3\pi }}{2}} \right]$

अब, $\sin (\theta  + \phi ) = \sin \theta .\cos \phi + \cos \theta .\sin \phi $

$ = \left( {\frac{{12}}{{13}}} \right)\,\left( {\frac{{ - 3}}{5}} \right) + \left( {\frac{5}{{13}}} \right)\,\left( {\frac{{ - 4}}{5}} \right)$

$= \frac{{ - 36}}{{65}} - \frac{{20}}{{65}}$

$ = \frac{{ - 56}}{{65}}$.

$\therefore$  $\cos (P - Q) = \cos P\cos Q + \sin P\sin Q$

$ = \frac{1}{7}.\frac{{13}}{{14}} + \frac{{\sqrt {48} }}{7}.\frac{{\sqrt {27} }}{{14}} $

$= \frac{{13 + 36}}{{98}} = \frac{1}{2} = \cos {60^o}$

==> $P - Q = {60^o}$.

$ = \tan ({45^o} + {10^o}) = \tan {55^o}$.

$\therefore $$\tan \,{225^o} = \frac{{\tan \,\,{{100}^o} + \tan \,\,{{125}^o}}}{{1 - \tan \,{{100}^o}\,\tan \,{{125}^o}}}$

अर्थात्, $1 = \frac{{\tan \,{{100}^o} + \tan \,\,{{125}^o}}}{{1 - \tan \,\,{{100}^o}\,\tan \,\,{{125}^o}}}$

अर्थात्, $\tan {100^o} + \tan {125^o} + \tan {100^o}\tan {125^o} = 1.$

$\therefore $ $5\,\,\cos A\,\,\cos B + 5\,\,\sin A\,\,\sin B = 3$…..$(i)$

द्वितीय सम्बन्ध से, $\sin A\sin B = 2\cos A\cos B$.....$(ii)$

$\therefore $ $\cos A\cos B = \frac{1}{5}$

एवं $5\,\left( {\frac{1}{2} + 1} \right)\,\sin A\,\,\sin B = 3$.

$ = \frac{{\sin 70^\circ + \sin 50^\circ }}{{\sin 20^\circ + \sin 40^\circ }} $

$= \frac{{2\sin 60^\circ \cos 10^\circ }}{{2\sin 30^\circ \cos ( - 10^\circ )}}$ 

$ = \frac{{\sin \,\,{{60}^o}}}{{\sin \,\,{{30}^o}}} = \frac{{\sqrt 3 }}{2}.\frac{2}{1} = \sqrt 3 $.

$=  \tan \,\left( {{{45}^o} + {9^o}} \right) = \tan {54^o}$.

$\frac{{\cos \,\,{{17}^o} + \sin \,\,{{17}^o}}}{{\cos \,\,{{17}^o} - \sin \,\,{{17}^o}}}$

$ = \frac{{1 + \tan \,\,{{17}^o}}}{{1 - \tan \,\,{{17}^o}}} $

$= \frac{{\tan \,\,{{45}^o} + \tan \,\,{{17}^o}}}{{1 - \tan \,\,{{45}^o}\tan \,\,{{17}^o}}} = \tan \,\,{62^o}$.

$ = (\cos \,\,{52^o} + \cos \,\,{172^o}) + \cos \,\,{68^o}$

$ = 2\,\,\cos \,\,{112^o}\,\cos \,\,{60^o} + \cos \,\,{68^o}$

$ = \cos \,\,{112^o} + \cos \,\,{68^o} = 2\,\,\cos \,\,({90^o})\,\,\cos \,\,{22^o} = 0$.

$ = (\cos \,\,{12^o} + \cos \,\,{132^o}) + (\cos \,\,{84^o} + \cos \,\,{156^o})$

$ = 2\,\,\cos {72^o}\cos \,{\kern 1pt} {60^o} + 2\cos \,\,{120^o}\cos \,\,{36^o}$

$ = 2\,\left[ {\cos \,\,{{72}^o} \times \frac{1}{2} - \frac{1}{2} \times \cos \,\,{{36}^o}} \right]$

$ = [\cos \,\,{72^o} - \cos \,{36^o}]$

$ = \left[ {\frac{{\sqrt 5  - 1}}{4} - \frac{{\sqrt 5  + 1}}{4}} \right] = \frac{{ - 1}}{2}$.

$ \Rightarrow \,\,\tan \frac{{6\pi }}{{15}} - \tan \frac{\pi }{{15}} = \sqrt 3  + \sqrt 3 \,\tan \frac{{6\pi }}{{15}}\tan \frac{\pi }{{15}}$

$ \Rightarrow \,\,\tan \frac{{6\pi }}{{15}} - \tan \frac{\pi }{{15}} - \sqrt 3 \,\tan \frac{{6\pi }}{{15}}\tan \frac{\pi }{{15}} = \sqrt 3 $.

$= \frac{{2\,\sin \,(A + B)\,\sin \,(A - B)}}{{\sin \,2A - \sin \,2B}}$

$ = \frac{{2\,\sin \,(A + B)\,\sin \,(A - B)}}{{2\,\cos \,(A + B)\,\sin \,(A - B)}} $

$= \tan \,(A + B)$.

लेकिन $\cos \,(A + B) = \cos \,A\,\cos B - \sin A\,\sin \,B$

$ \Rightarrow \,\,\alpha  = 1,\,\,\beta  =  - 1.$

$ = \frac{1}{2}\{ \cos {30^o}\cos {23^o} - \sin {30^o}\sin {23^o}\} $

$ = \frac{1}{2}\,\cos \,({30^o} + {23^o}) $

$= \frac{1}{2}\,\cos \,{53^o}.$

$= \cos \,{15^o} = \cos \,\,({45^o} - {30^o})$

$ = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$.

$\therefore \,\,{\cos ^2}{48^o} - {\sin ^2}{12^o} = \cos \,\,{60^o}\,.\,\cos \,\,{36^o}$

$ = \frac{1}{2}\,\left( {\frac{{\sqrt 5 + 1}}{4}} \right) = \frac{{\sqrt 5 + 1}}{8}.$

$ = - 2\,\,\cos \,\,{60^o}\sin \,\,{10^o} + \sin \,\,{10^o}$

$ = \,\sin \,{10^o}\,(1 - 2\,\,\cos \,\,{60^o}) = 0.$

$ = 2\frac{{(2\tan B + \cot B - \tan B)}}{{1 + (2\,\tan B + \cot B)\,\tan B}} $

$= 2\,\frac{{\tan B + \cot B}}{{2\,(1 + {{\tan }^2}B)}}$

$ = \frac{{\cot B\,({{\tan }^2}B + 1)}}{{(1 + {{\tan }^2}B)}} $

$= \cot B$.

$ = \frac{1}{{\sqrt {10} }}\sqrt {1 - \frac{1}{5}}  + \frac{1}{{\sqrt 5 }}\,\sqrt {1 - \frac{1}{{10}}} $

$ = \frac{1}{{\sqrt {10} }}\sqrt {\frac{4}{5}}  + \frac{1}{{\sqrt 5 }}\sqrt {\frac{9}{{10}}} $

$= \frac{1}{{\sqrt {50} }}(2 + 3) = \frac{5}{{\sqrt {50} }} = \frac{1}{{\sqrt 2 }}$

$ \Rightarrow \,\,\sin \,(A + B) = \sin \frac{\pi }{4}$

अत: $A + B = \frac{\pi }{4}$

$ = \sin \theta .\sin \theta + \cos \theta .\cos \theta = 1$.

$ = ({\sec ^2}A - {\tan ^2}A) + \sec A + \tan A - \sec A$$ + \tan A - 1 - 2\tan A = 0$

                                           $( \because {\sec ^2}A - {\tan ^2}A = 1)$

$ \Rightarrow \frac{x}{1} = \frac{y}{{ - 2}} = \frac{z}{{ - 2}} = \lambda $(माना)

$ \Rightarrow x = \lambda ,\;y =  - 2\lambda ,\,z =  - 2\lambda $

$\therefore xy + yz + zx =  - 2{\lambda ^2} + 4{\lambda ^2} - 2{\lambda ^2} = 0$.

$ = \tan A + \cot A - \tan A - \cot A = 0$.

$= \tan (90^\circ - 45^\circ - \theta )\cot (45^\circ - \theta )$

$ = \tan (45^\circ - \theta )\cot (45^\circ - \theta ) = 1$.

$ = - \sin \theta \sin \theta \frac{1}{{{{\sin }^2}\theta }} = - 1$.

$= \frac{{1 + \tan \theta }}{{1 - \tan \theta }} - \frac{{1 - \tan \theta }}{{1 + \tan \theta }}$

$ = \frac{{4\tan \theta }}{{1 - {{\tan }^2}\theta }} $

$= 2\left( {\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right) $

$= 2\tan 2\theta $.

$=\cos A - \cos A + \cos A - \cos A = 0$.

$= \cos 15^\circ - \sin 15^\circ $

$= \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} - \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }} $

$= \frac{2}{{2\sqrt 2 }} = \frac{1}{{\sqrt 2 }}$.

$\sin 15^\circ + \cos (90^\circ + 15^\circ ) = \sin 15^\circ - \sin 15^\circ = 0$.

==> $\tan B = \frac{{\tan A + \tan C}}{{1 - \tan A\tan C}}$

==> $\tan A\tan B\tan C = \tan B - \tan A - \tan C$.

$⇒$  $ - \cos (B + C) = \cos B\cos C$

$⇒$  $ - [\cos B\cos C - \sin B\sin C] = \cos B\cos C$ 

$⇒$  $\sin B\sin C = 2\cos B\cos C$ 

$⇒$  $\tan B\tan C = 2$.

$\Rightarrow {(\sin x - 1)^2} = 0 \Rightarrow \sin x = 1$ 

$\therefore {\sin ^n}x + {\rm{cose}}{{\rm{c}}^n}x = 1 + 1 = 2$ .

$= \sqrt {1 + {{\cot }^2}\alpha + 2\cot \alpha } = \,\,|1 + \cot \alpha |$

लेकिन $\frac{{3\pi }}{4} < \alpha < \pi \Rightarrow \cot \alpha < - 1$

$\Rightarrow 1 + \cot \alpha < 0$

अतः  $|1 + \cot \alpha | = - (1 + \cot \alpha )$.

अत: तार की लम्बाई $=   10\pi $.

अत: अभीष्ट कोण $=$ चाप$/$त्रिज्या $ = \frac{{10\pi }}{{50}} = \frac{\pi }{5}$ रेडियन

$= \sqrt {\frac{{1 + \cos {{30}^o}}}{2}} $ .$\left( {\,\,\,\,\because \cos {{15}^o} > 0} \right)$

$\Rightarrow A = {30^o}$

==> $\tan 3A = \tan {90^o} = \infty $.

$ = - \cos \frac{\pi }{4} + \sin \frac{\pi }{4} = - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = 0$.

$ = - \tan [(2 \times 360^\circ + 225^\circ )]$

$ = - \tan [225^\circ ] = - \tan 45^\circ = - 1$.

$= 2\cos 85^\circ \cos 45^\circ > 0$.

$\Rightarrow \frac{x}{{4\sqrt 2 }} = \sqrt 2 $

$\Rightarrow x = 8$.

$ = \sin 18^\circ .\cos 36^\circ = \frac{{\sqrt 5 - 1}}{4}.\frac{{\sqrt 5 + 1}}{4} = \frac{1}{4}$.

                        $ + (\cos 89^\circ + \cos 91^\circ ) + \cos 90^\circ + \cos 180^\circ = - 1$.

$\sin 210^\circ  =  - \sin 30^\circ ,\;\sin 360^\circ  = \sin 180^\circ  = 0$ इत्यादि।

इसलिए $\frac{{\cot 54^\circ }}{{\tan 36^\circ }} + \frac{{\tan 20^\circ }}{{\cot 70^\circ }}$

$ = \frac{{\cot 54^\circ }}{{\tan (90^\circ  - 54^\circ )}} + \frac{{\tan 20^\circ }}{{\cot (90^\circ  - 20^\circ )}}$

$= \frac{{\cot 54^\circ }}{{\cot 54^\circ }} + \frac{{\tan 20^\circ }}{{\tan 20^\circ }} = 1 + 1 = 2$.

इसलिए $\cos 1^\circ .\cos 2^\circ .\cos 3^\circ ...\cos 179^\circ  = 0$.

$\therefore$ $\frac{{{x^2} + {y^2} + 1}}{{2x}} \le 1$

$\Rightarrow$ ${x^2} + {y^2} - 2x + 1 \le 0$

$\Rightarrow$ ${(x - 1)^2} + {y^2} \le 0$

यह सम्भव है यदि $x = 1$ एवं $y = 0$,

अर्थात् यह $y$ के मान पर भी निर्भर है।

अत: विकल्प  $(d)$ सही है।  

$ \Rightarrow \sin {\theta _1} = \sin {\theta _2} = \sin {\theta _3} = 1$,       $( \because - 1 \le \sin x \le 1)$

$ \Rightarrow {\theta _1} = {\theta _2} = {\theta _3} = \frac{\pi }{2}$

$\Rightarrow \cos {\theta _1} + \cos {\theta _2} + \cos {\theta _3} = 0$.

$ = \tan \alpha \tan \beta \tan \gamma $          ...$(i)$

माना $x = (\sec \alpha  - \tan \alpha )(\sec \beta  - \tan \beta )(\sec \gamma  - \tan \gamma )$ ...$(ii)$

समी. $(i)$ व $(ii)$ को गुणा करने पर,

$({\sec ^2}\alpha  - {\tan ^2}\alpha )({\sec ^2}\beta  - {\tan ^2}\beta )({\sec ^2}\gamma  - {\tan ^2}\gamma )$

$ = x.(\tan \alpha \tan \beta \tan \gamma )$

$ \Rightarrow x = \frac{1}{{\tan \alpha \tan \beta \tan \gamma }}$,

$\therefore x = \cot \alpha \cot \beta \cot \gamma $

$(1 - {\sin ^2}A)(1 - {\sin ^2}B)(1 - {\sin ^2}C)$   

$ = {(1 - \sin A)^2}{(1 - \sin B)^2}{(1 - \sin C)^2}$

$\Rightarrow$ $(1 - \sin A)(1 - \sin B)(1 - \sin C) =  \pm \cos A\cos B\cos C$

इसी प्रकार,  $(1 + \sin A)(1 + \sin B)(1 + \sin C) =  \pm \cos A\cos B\cos C$.

$ \Rightarrow \,\,{\sin ^6}\theta  + {\cos ^6}\theta  + 3\,{\sin ^2}\theta \,{\cos ^2}\theta  = 1$

तथा ${\sin ^4}\theta  + {\cos ^4}\theta  + 2\,{\sin ^2}\theta \,{\cos ^2}\theta  = 1$

$\therefore$ $2\,\,({\sin ^6}\theta  + {\cos ^6}\theta ) - 3\,({\sin ^4}\theta  + {\cos ^4}\theta ) + 1 = 0$.

$ = {({\sin ^2}\theta  + {\cos ^2}\theta )^3} - 3{\sin ^2}\theta {\cos ^2}\theta  + 3{\sin ^2}\theta {\cos ^2}\theta  = 1.$

ट्रिक : $\theta  = {0^o}$ रखने पर, व्यंजक का मान $1$ है।

पुन: $\theta  = {45^o}$ रखने पर मान $1$ आता है,

इसका अर्थ है कि व्यंजक $\theta$ से स्वतंत्र व $1$ के बराबर है।  

वर्ग करके योग कीजिए।

एवं $a\,\sin \,\theta  - b\,\cos \theta  = n.$

वर्ग करके जोड़ने पर

${(a\,\,\cos \theta  + b\,\sin \theta )^2} + {(a\,\sin \theta  - b\,\cos \theta )^2} = {m^2} + {n^2}$

$ \Rightarrow \,\,{a^2}({\cos ^2}\theta  + {\sin ^2}\theta ) + {b^2}({\cos ^2}\theta  + {\sin ^2}\theta )$

$ + 2ab\,(\cos \theta \,\sin \theta  - \sin \theta \,\cos \theta ) = {m^2} + {n^2}$

अत:, ${a^2} + {b^2} = {m^2} + {n^2}.$

ट्रिक : यहाँ हम अनुमान लगा सकते हैं कि ${a^2} + {b^2}$ का मान $\theta$ से स्वतंत्र है।

अत: $\theta$  का कोई भी मान रखने पर अर्थात्  $\frac{\pi }{2},$

$b = m$ व $a = n.$ अत: ${a^2} + {b^2} = {m^2} + {n^2}$ 

($\theta$ के अन्य मान के लिए परीक्षण करें).

$ \Rightarrow \,{\tan ^2}A + {\cot ^2}A + 2\,\tan A\,\,\cot A = 16$

$ \Rightarrow \,{\tan ^2}A + {\cot ^2}A = 14\,\, $

$\Rightarrow \,{\tan ^4}A + {\cot ^4}A + 2 = 196$

$ \Rightarrow \,{\tan ^4}A + {\cot ^4}A = 194.$

$ = \frac{{2\,\sin \theta }}{{{{(1 + \tan \,\theta )}^2}}}\,\{ \tan \,\theta \,(1 - \tan \,\theta ) + {\sec ^2}\theta \} $

$ = \frac{{2\,\sin \theta }}{{{{(1 + \tan \,\theta )}^2}}}\,\{ \tan \,\theta \, - {\tan ^2}\,\theta  + 1 + {\tan ^2}\theta \} $

$ = \frac{{2\,\sin \theta }}{{1 + \tan \theta }}$.

$ = \frac{{2\,\cos \,2x}}{{\sin \,2x}} = 2\,\,\cot \,\,2x.$

$\Rightarrow \,x + \frac{1}{x} = 2\,\cos \theta $

हम जानते हैं कि, ${x^2} + \frac{1}{{{x^2}}} = {\left( {x + \frac{1}{x}} \right)^2} - 2$

$ = {(2\cos \theta )^2} - 2 = 4\,{\cos ^2}\theta  - 2 = 2\,\cos \,\,2\theta $

$\therefore \,\,\frac{1}{2}\,\left( {{x^2} + \frac{1}{{{x^2}}}} \right) $

$= \frac{1}{2} \times 2\,\cos \,2\theta  = \cos \,2\theta $

${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}\alpha $.

${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}\alpha  - 2$,

${x^2} + \frac{1}{{{x^2}}} = 2(2{\cos ^2}\alpha  - 1) = 2\cos 2\alpha $

इसी प्रकार  ${x^n} + \frac{1}{{{x^n}}} = 2\cos \,n\alpha $.

$ \Rightarrow \,x + \frac{1}{x} = \sec \theta + \tan \theta + \frac{1}{{\sec \theta + \tan \theta }}$

$ = \sec \theta + \tan \theta + \sec \theta - \tan \theta = 2\sec \theta $

$\sec \theta  - \tan \theta  = \frac{1}{p}$…..$(ii)$

$(ii)$ को $(i)$ से घटाने पर,

$2\tan \theta  = p - \frac{1}{p}$

$ \Rightarrow \,\tan \theta  = \frac{{{p^2} - 1}}{{2p}}$.

$ = \frac{{\sin \theta \,.\,\sin \theta }}{{\,\sin \theta - \cos \theta }} + \frac{{\cos \theta \,.\cos \theta }}{{\cos \theta - \sin \theta }}$

$ = \frac{{({{\cos }^2}\theta - {{\sin }^2}\theta )}}{{(\cos \theta - \sin \theta )}} = \cos \theta + \sin \theta $.

अत: योग भी धनात्मक होगा, परन्तु $\frac{\pi }{2} < \theta < \pi ,$ के लिए योग

$\frac{{1 - \sin \theta  + 1 + \sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }} = \frac{2}{{\cos \theta }};$ जो कि ऋणात्मक है।

($\because$ $\cos \theta $ द्वितीय चतुर्थांश में ऋणात्मक होता है)

अत: अभीष्ट धनात्मक मान

$ = \frac{{ - 2}}{{\cos \theta }} =  - 2\,\sec \theta ,\,\left( {\frac{\pi }{2} < \theta  < \pi } \right)$.

$ \Rightarrow \,\,\sin A\, =  \pm \,\frac{{\tan A}}{{\sqrt {1 + {{\tan }^2}A} }} =  \pm \frac{{ - 4/3}}{{\sqrt {1 + 16/9} }} = \frac{4}{5}$

($\because$ $A$ द्वितीय चतुर्थांश में है)

एवं $\cos \,A =  - \frac{3}{5}$. अत: $2\cot A - 5\cos A + \sin A$

$ = 2\,\left( { - \frac{3}{4}} \right) - 5\,\left( { - \frac{3}{5}} \right) + \frac{4}{5} = \frac{{23}}{{10}}$.

अत: $\cos \theta  = $ धनात्मक.

अब $1 + {\tan ^2}\theta  = {\sec ^2}\theta $

$\Rightarrow 1 + \frac{1}{{10}} = {\sec ^2}\theta $

$ \Rightarrow {\sec ^2}\theta  = \frac{{11}}{{10}}$

$\Rightarrow \cos \theta  = \sqrt {\left( {\frac{{10}}{{11}}} \right)} $..

एवं $\cos (\alpha  + \beta ) = \frac{1}{2} \Rightarrow \alpha  + \beta  = 60^\circ $…..$(ii)$

$(i)$ व $(ii)$ को हल करने पर $\alpha  = 45^\circ $ एवं $\beta  = 15^\circ $.

ट्रिक : विद्यार्थी इस प्रकार के प्रष्नों में दिये गये प्रतिबंध को विकल्पों के मान द्वारा सन्तुष्ट कर सकते हैं,

यहाँ  $\alpha  = 45^\circ, \beta  = 15^\circ $ दोनों प्रतिबन्धों को सन्तुष्ट करते हैं ।

$ \Rightarrow \cos \theta  = \sqrt {1 - \frac{{16}}{{25}}}  =  \pm \frac{3}{5}$

$\cos \frac{\theta }{2} =  \pm \sqrt {\frac{{1 + \cos \theta }}{2}} $

$= \sqrt {\frac{{1 - 3/5}}{2}}  =  \pm \sqrt {\frac{1}{5}} $

लेकिन $\cos \frac{\theta }{2} =  - \frac{1}{{\sqrt 5 }}$

चूँकि $\frac{\theta }{2}$ द्वितीय चतुर्थांष  में है।

अत: $\cos \frac{\theta }{2} =  - \frac{1}{{\sqrt 5 }}$.

$ \Rightarrow $$\sin \theta  = \sin 225^\circ  $

$\Rightarrow \theta  = 225^\circ $

चूँकि तृतीय चतुर्थांश में $\sin \theta $ ऋणात्मक व $\tan \theta $ धनात्मक है।

$= \sqrt {1 - {{\left( {\frac{{ - 24}}{{25}}} \right)}^2}} = \frac{7}{{25}}$

$ \Rightarrow \tan x = \frac{{\sin x}}{{\cos x}} = \frac{{ - 24}}{7}$.

$\therefore \sin \theta = \frac{4}{{\sqrt {41} }}$ एवं 

$\cos \theta = \frac{5}{{\sqrt {41} }}$

$\frac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }}$

$= \frac{{5 \times \frac{4}{{\sqrt {41} }} - 3 \times \frac{5}{{\sqrt {41} }}}}{{5 \times \frac{4}{{\sqrt {41} }} + 2 \times \frac{5}{{\sqrt {41} }}}}$

$\frac{{20 - 15}}{{20 + 10}} = \frac{5}{{30}} = \frac{1}{6}$.

$\Rightarrow {\rm{cosec}}\,A - \cot A = \frac{2}{{11}}$

अत: $2\cot A = \frac{{117}}{{22}}$

$\Rightarrow \tan A = \frac{{44}}{{117}}$.

.

$\Rightarrow \cos \theta = \frac{{ - 7}}{{25}},\,\tan \theta = \frac{{ - 24}}{7}$

$\therefore$  $\sec \theta  + \tan \theta  = \frac{{ - 25}}{7} + \frac{{ - 24}}{7} =  - 7$

 दोनों पक्षों का वर्ग करने पर

$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 1$

$\therefore \sin \theta \cos \theta = 0$.

$ = \frac{{\sin \theta + \cos \theta }}{{\sin \theta .\cos \theta }}.2\sin \theta \cos \theta = 2m$.

$  = {(\sin \theta  + {\rm{cosec}}\theta )^2} - 2\sin \theta {\rm{ cosec}}\theta $

$ = {(2)^2} - 2 = 4 - 2 = 2,$

चूँकि $(\sin \theta  + {\rm{cosec}}\theta ) = 2.$

$ \Rightarrow $${\sin ^2}\theta  + 1 = 2\sin \theta $

$ \Rightarrow $${\sin ^2}\theta  - 2\sin \theta  + 1 = 0$

$ \Rightarrow $${(\sin \theta  - 1)^2} = 0 \Rightarrow \sin \theta  = 1$

${\sin ^{10}}\theta  + {\rm{cose}}{{\rm{c}}^{10}}\theta $ का अभीष्ट मान

$ = {(1)^{10}} + \frac{1}{{{{(1)}^{10}}}} = 2$.

$ = (\tan 1^\circ \tan 89^\circ )(\tan 2^\circ \tan 88^\circ )......$$ = 1 \times 1 \times 1.... = 1.$

चूँकि $\sin \theta $ का मान $\left[ {0 \to \frac{\pi }{2}} \right]$ में वर्धमान है।

and $\tan 2=-2.185$

Clearly, $\tan 1 > \tan 2$

त्रिज्या $= 20$ सेमी

अत: वृत्तीय तार की लम्बाई $= 14\pi$ सेमी

$\therefore $ अभीष्ट कोण  $ = \frac{{{\rm{arc}}}}{{{\rm{radius}}}} = \frac{{14\pi }}{{12}} = \frac{{7\pi }}{6}$

$ = \frac{7}{6}\pi .\frac{{{{180}^o}}}{\pi } = {210^o}$

हम जानते हैं कि कोण $= \frac{{{\rm{arc}}}}{{{\rm{radius}}}} = \frac{1}{3}$ रेडियन

$ = \frac{3}{8}\,.\,2\sin x\sin 3x - \frac{1}{8}\,.\,2\,{\sin ^2}3x$

$ = \frac{3}{8}(\cos 2x - \cos 4x) - \frac{1}{8}(1 - \cos 6x)$

$ = - \frac{1}{8} + \frac{3}{8}\cos 2x - \frac{3}{8}\cos 4x + \frac{1}{8}\cos 6x$.....$(i)$

तथा $\sum\limits_{m = 0}^n {{c_m}\cos mx} = {c_0} + {c_1}\cos x + {c_2}\cos 2x$

$ + {c_3}\cos 3x + ...... + {c_n}\cos nx$..…$(ii)$

समी. $(i)$ व $(ii)$ के दोनों पक्षों की तुलना करने पर $n = 6$.

==> $\frac{{{{(1 - \cos 2A)}^2}}}{{4a}} + \frac{{{{(1 + \cos 2A)}^2}}}{{4b}} = \frac{1}{{a + b}}$

==> $b(a + b)(1 - 2\cos 2A + {\cos ^2}2A)$

$ + a(a + b)(1 + 2\cos 2A + {\cos ^2}2A) = 4ab$

==>$\{ b(a + b) + a(a + b)\} {\cos ^2}2A + 2(a + b)(a - b)\cos 2A$

$ + a(a + b) + b(a + b) - 4ab = 0$

==> ${(a + b)^2}{\cos ^2}2A + 2(a + b)(a - b)\cos 2A + {(a - b)^2} = 0$

==> ${\{ (a + b)\cos 2A + (a - b)\} ^2} = 0$या $\cos 2A = \frac{{b - a}}{{b + a}}$

अत: $\frac{{{{\sin }^8}A}}{{{a^3}}} + \frac{{{{\cos }^8}A}}{{{b^3}}} = \frac{{{{(1 - \cos 2A)}^4}}}{{16{a^3}}} + \frac{{{{(1 + \cos 2A)}^4}}}{{16{b^3}}}$

$ = \frac{1}{{16{a^3}}}{\left[ {1 - \frac{{b - a}}{{b + a}}} \right]^4} + \frac{1}{{16{b^3}}}{\left[ {1 + \frac{{b - a}}{{b + a}}} \right]^4}$

$ = \frac{{16{a^4}}}{{16{a^3}{{(b + a)}^4}}} + \frac{{16{b^4}}}{{16{b^3}{{(b + a)}^4}}}$

$ = \frac{1}{{{{(b + a)}^4}}}(a + b) = \frac{1}{{{{(a + b)}^3}}}$

ट्रिक : $A = {90^o}$ रखने पर,

$\frac{{{{\sin }^4}A}}{a} + \frac{{{{\cos }^4}A}}{b} = \frac{1}{{a + b}}$==>$\frac{1}{a} = \frac{1}{{a + b}} \Rightarrow b = 0$

$\therefore \,\,\,\,\frac{{{{\sin }^8}A}}{{{a^3}}} + \frac{{{{\cos }^8}A}}{{{b^3}}} = \frac{1}{{{a^3}}}$ जो कि विकल्प $(a)$ द्वारा दिया जाता है।

नोट : विद्यार्थी  इस प्रश्न का $A$ के अन्य मानों के लिए भी निरीक्षण करें।

$\therefore $ अत: $4{x^2} - 16x + 15 < 0$ का पूर्णांक हल  $x = 2$ है।

अत: $\tan \alpha  = 2$ यह दिया गया है कि $\cos \beta  = \tan {45^o} = 1$

$\therefore \,\,\sin \,(\alpha  + \beta )\,\sin \,(\alpha  - \beta ) = {\sin ^2}\alpha  - {\sin ^2}\beta $

$ = \frac{1}{{1 + {{\cot }^2}\alpha }} - (1 - {\cos ^2}\beta ) = \frac{1}{{1 + \frac{1}{4}}} - 0 = \frac{4}{5}$.

$\therefore$ $\tan \alpha  + \tan \beta  =  - p,$ $\tan \alpha \tan \beta  = q$

$\therefore$ $\tan \,(\alpha  + \beta ) = \frac{{\tan \alpha  + \tan \beta }}{{1 - \tan \alpha \tan \beta }} = \frac{p}{{q - 1}}$,

जो $(b)$ में दिया गया है।

एवं जब $\tan \,(\alpha  + \beta ) = \frac{p}{{q - 1}}$.

तब (a) में व्यंजक का L.H.S.

$ = {\cos ^2}(\alpha  + \beta )\,\,[{\tan ^2}(\alpha  + \beta ) + p\tan \,(\alpha  + \beta ) + q]$

$ = \frac{1}{{1 + {{\tan }^2}(\alpha  + \beta )}}\,\left[ {\frac{{{p^2}}}{{{{(q - 1)}^2}}} + \frac{{{p^2}}}{{q - 1}} + q} \right]$

$ = \frac{{{{(q - 1)}^2}}}{{{{(q - 1)}^2} + {p^2}}}\left[ {\frac{{{p^2} + {p^2}(q - 1) + q\,{{(q - 1)}^2}}}{{{{(q - 1)}^2}}}} \right]$

$ = \frac{{q\,\left\{ {{p^2} + {{(q - 1)}^2}} \right\}}}{{{p^2} + {{(q - 1)}^2}}} $

$= q = (a)$ का $R.H.S.$

अर्थात् $(a)$ में दिया गया सम्बन्ध सन्तुष्ट होता हैं।

और $\sin \beta = 3/5 \Rightarrow \cos \beta = 4/5$

$\sin (\beta - \alpha ) = \sin \beta \cos \alpha - \sin \alpha \cos \beta $

$ = \frac{3}{5}.\frac{2}{{\sqrt 5 }} - \frac{1}{{\sqrt 5 }}.\frac{4}{5} = \frac{2}{{5\sqrt 5 }} = 0.1789$

अब  $\sin \frac{\pi }{4} = \frac{1}{{\sqrt 2 }} = 0.7071 = \sin \frac{{3\pi }}{4}$

चूँकि  $0 < 0.1789 < 0.7071$

$\therefore $$\sin 0 < \sin (\beta - \alpha ) < \sin \frac{\pi }{4}$

$\Rightarrow 0 < (\beta - \alpha ) < \frac{\pi }{4}$

तथा  $\sin \pi < \sin (\beta - \alpha ) < \sin \frac{{3\pi }}{4}$

$\therefore $$(\beta - \alpha ) \in [0,\,\pi /4]$ और $[3\pi /4,\,\pi ]$.

$= \frac{1}{4}\,\left[ {{{\left( {2{{\sin }^2}\frac{\pi }{8}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{{3\pi }}{8}} \right)}^2}} \right]$

$ + \frac{1}{4}\,\left[ {{{\left( {2{{\sin }^2}\frac{\pi }{8}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{{3\pi }}{8}} \right)}^2}} \right]$

= $\frac{1}{4}\,\left[ {{{\left( {1 - \cos \frac{\pi }{4}} \right)}^2} + {{\left( {1 - \cos \frac{{3\pi }}{4}} \right)}^2}} \right]$

$ + \frac{1}{4}\,\left[ {{{\left( {1 - \cos \frac{\pi }{4}} \right)}^2} + {{\left( {1 - \cos \frac{{3\pi }}{4}} \right)}^2}} \right]$

$=  \frac{1}{4}\,\left[ {{{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)}^2}} \right] + \frac{1}{4}\,\left[ {{{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)}^2}} \right]$

$= \frac{1}{4}(3)\, + \frac{1}{4}(3) = \frac{3}{2}$.

$\Rightarrow (b - a) = (c - a)(1 + {\tan ^2}x)$

$b{\sin ^2}y + a{\cos ^2}y = d \Rightarrow (a - b){\cos ^2}y = d - b$ 

$ \Rightarrow (a - b) = (d - b)(1 + {\tan ^2}y)$

$\therefore {\tan ^2}x = \frac{{b - c}}{{c - a}},\,\,{\tan ^2}y = \frac{{a - d}}{{d - b}}$

$\therefore \frac{{{{\tan }^2}x}}{{{{\tan }^2}y}} = \frac{{(b - c)(d - b)}}{{(c - a)(a - d)}}$…..$(i)$

लेकिन  $a\tan x = b\tan y,$

अर्थात् $\frac{{\tan x}}{{\tan y}} = \frac{b}{a}$…..$(ii)$ 

$(i)$ व   $(ii) $ से,

$\frac{{{b^2}}}{{{a^2}}} = \frac{{(b - c)(d - b)}}{{(c - a)(a - d)}}$

$ \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{(c - a)(a - d)}}{{(b - c)(d - b)}}$.

$(\therefore  (b) $ हल है $)$

$y - z = a({\cos ^2}x - {\sin ^2}x) + 4b\sin x\cos x$

$ - c({\cos ^2}x - {\sin ^2}x)$

$ = (a - c)\cos 2x + 2b\sin 2x$

$ = (a - c).\,\left( {\frac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} \right) + 2b.\left( {\frac{{2\tan x}}{{1 + {{\tan }^2}x}}} \right)$

$ = (a - c).\left\{ {\frac{{1 - 4{b^2}/{{(a - c)}^2}}}{{1 + 4{b^2}/{{(a - c)}^2}}}} \right\} + 2b.\left\{ {\frac{{2.2b/(a - c)}}{{1 + 4{b^2}/{{(a - c)}^2}}}} \right\}$

चूँकि  $\tan x = \frac{{2b}}{{(a - c)}}$,

$\therefore y - z = \frac{{(a - c).\{ {{(a - c)}^2} - 4{b^2}\} + 8{b^2}(a - c)}}{{{{(a - c)}^2} + 4{b^2}}}$

$ = \frac{{(a - c){{(a - c)}^2} + 4{b^2}}}{{\{ {{(a - c)}^2} + 4{b^2}\} }} = (a - c)$

$ \Rightarrow y \ne z\,,\,(a \ne c)$

$ = \sin (\theta - \beta )\cos (\theta - \alpha ) - \cos (\theta - \beta )\sin (\theta - \alpha )$

$ = ba - \sqrt {1 - {b^2}} \sqrt {1 - {a^2}} $

एवं  $\cos (\alpha - \beta ) = \cos (\theta - \beta - \overline {\theta - \alpha } )$

$ = \cos (\theta - \beta )\cos (\theta - \alpha ) + \sin (\theta - \beta )\sin (\theta - \alpha )$

$ = a\sqrt {1 - {b^2}} + b\sqrt {1 - {a^2}} $

$\therefore $ अतः दिया गया व्यंजक ${\cos ^2}(\alpha - \beta ) + 2ab\sin (\alpha - \beta )$

$ = (a\sqrt {1 - {b^2}} + b\sqrt {1 - {a^2}{)^2}} + 2ab\{ ab - \sqrt {1 - {a^2}} \sqrt {1 - {b^2}} \} $

$ = {a^2} + {b^2}$.

ट्रिक :  $\alpha = 30^\circ ,\beta = 60^\circ ,$$\theta = 90^\circ $ रखने पर

$a = \frac{1}{2},b = \frac{1}{2}$

$\therefore {\cos ^2}(\alpha - \beta ) + 2ab\sin (\alpha - \beta ) = \frac{3}{4} + \frac{1}{2} \times \left( { - \frac{1}{2}} \right) = \frac{1}{2}$

जो कि विकल्प $(c)$ द्वारा दिया जाता है। 

$ \Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$

अब समान्तर माध्य $\ge$ गुणोत्तर माध्य

$ \Rightarrow \frac{{\tan A + \tan B + \tan C}}{3} \ge {(\tan A\tan B\tan C)^{1/3}}$

$ \Rightarrow \left( {\frac{{\tan A\tan B\tan C}}{3}} \right) \ge {(\tan A\tan B\tan C)^{1/3}}$

$ \Rightarrow {(\tan A\tan B\tan C)^{2/3}} \ge 3$

$ \Rightarrow {\left( {\frac{1}{K}} \right)^{2/3}} \ge 3$

$\Rightarrow \frac{1}{K} \ge {3^{3/2}} $

$\Rightarrow K \le \frac{1}{{3\sqrt 3 }}$.

$= \frac{{2\sin A\cos A}}{{2{{\cos }^2}A}} = \frac{{\sin 2A}}{{1 + \cos 2A}}$

$A = 7{\frac{1}{2}^o}$ रखने पर,

$ \Rightarrow \tan 7{\frac{1}{2}^o} = \frac{{\sin {{15}^o}}}{{1 + \cos {{15}^o}}}$

सरल करने पर, $\tan 7{\frac{1}{2}^o} = \sqrt 6  - \sqrt 3  + \sqrt 2  - 2$

$\Rightarrow$ $\cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  =  - \sqrt {1 - {{\left( {\frac{{336}}{{625}}} \right)}^2}} $

                                               [ $\because \alpha $, $II$  चतुर्थांश में है]

अब $\cos \left( {\frac{\alpha }{2}} \right) =  - \sqrt {\frac{{1 + \cos \alpha }}{2}}  =  - \frac{7}{{25}}$     

                                                [ $\because \frac{\alpha }{2}$  $III$ चतुर्थांश में है]

$\therefore \,\,\,\sin \left( {\frac{\alpha }{4}} \right) =  + \sqrt {\frac{{1 - \cos (\alpha /2)}}{2}}  = \sqrt {\frac{{1 + \frac{7}{{25}}}}{2}}  = \frac{4}{5}$

                                                 [ $\because \frac{\alpha }{4}$  दूसरे चतुर्थांश में है]

$ = {\cos ^4}\frac{\pi }{8} + {\cos ^4}\frac{{3\pi }}{8} + {\cos ^4}\frac{{3\pi }}{8} + {\cos ^4}\frac{\pi }{8}$

$ = 2\left( {{{\cos }^4}\frac{\pi }{8} + {{\cos }^4}\frac{{3\pi }}{8}} \right)$

$ = 2\left[ {{{\left( {{{\cos }^2}\frac{\pi }{8} + {{\cos }^2}\frac{{3\pi }}{8}} \right)}^2} - 2{{\cos }^2}\frac{\pi }{8}{{\cos }^2}\frac{{3\pi }}{8}} \right]$

$ = 2\left[ {1 - \frac{1}{2}\left( {2{{\cos }^2}\frac{\pi }{8}} \right)\,\left( {2{{\cos }^2}\frac{{3\pi }}{8}} \right)} \right]$ 

$ = 2 - \left( {1 + \cos \frac{\pi }{4}} \right)\,\left( {1 + \cos \frac{{3\pi }}{4}} \right)$

$ = 2 - \left( {1 + \cos \frac{\pi }{4}} \right)\,\left( {1 - \cos \frac{\pi }{4}} \right)$

$ = 2 - \left( {1 - {{\cos }^2}\frac{\pi }{4}} \right) = 2 - \left( {1 - \frac{1}{2}} \right) $

$= 2 - \frac{1}{2} = \frac{3}{2}$.

$ = {\cos ^2}(A - B) + {\cos ^2}B$

$ - \cos \,(A - B)\,\left\{ {\cos (A - B) + \cos (A + B)} \right\}$

$ = {\cos ^2}B - \cos \,(A - B)\,\,\cos \,\,(A + B)$

$ = {\cos ^2}B - ({\cos ^2}A - {\sin ^2}B) = 1 - {\cos ^2}A$

अत:  $A$ पर निर्भर है।

ट्रिक :  $A$ के दो भिन्न मान रखने पर,

माना $A = {90^o},$ तब व्यंजक का मान ${\sin ^2}B + {\cos ^2}B = 1$

अब $A = {0^o}$ रखने पर मान ${\cos ^2}B + {\cos ^2}B - 2\,\,{\cos ^2}B = 0$ है

इसका अर्थ है कि व्यंजक, $A$ के भिन्न मानों के लिए भिन्न है अर्थात् यह $A$ पर निर्भर है।

इसी प्रकार $B = {90^o}$ के लिए व्यंजक ${\sin ^2}A + 0 - 0 = {\sin ^2}A$

व $B\,\, = {0^o}$ पर  ${\cos ^2}A + 1 - 2{\cos ^2}A = {\sin ^2}A$है।

अत: व्यंजक, $B$ के भिन्न मानों के लिए समान मान देता है

इसलिए यह $B$ पर निर्भर नहीं करता है।

एवं $\sin \,(\theta  + \beta ) = b$…..$(ii)$

अब, $\cos \,(\theta  + \alpha ) = \sqrt {1 - {a^2}} \, \Rightarrow \,\,\theta  + \alpha  = {\cos ^{ - 1}}\sqrt {1 - {a^2}} $

एवं $\alpha \, - \beta  = (\theta  + \alpha ) - (\theta  + \beta )$

$ = \,\,{\cos ^{ - 1}}\sqrt {1 - {a^2}}  - {\cos ^{ - 1}}\sqrt {1 - {b^2}} $

$ \Rightarrow \,\,\alpha  - \beta  = {\cos ^{ - 1}}(\sqrt {1 - {a^2}} \,\sqrt {1 - {b^2}}  + ab)$

$ \Rightarrow \,\,\cos \,(\alpha  - \beta ) = \sqrt {1 - {a^2}} \,\sqrt {1 - {b^2}}  + ab$

अब, $\cos \,\,2\,(\alpha  - \beta ) - 4ab\,\,\cos \,(\alpha  - \beta )$

$ = 2\,\,{\cos ^2}\,(\alpha  - \beta ) - 1 - 4ab\,\,\cos \,(\alpha  - \beta )$

$ = 2\,{\left( {\sqrt {1 - {a^2}} \sqrt {1 - {b^2}}  + ab} \right)^2}$

$ - 4ab\,\left( {\sqrt {1 - {a^2}} \sqrt {1 - {b^2}}  + ab} \right) - 1$

$ = 2\,\{ (1 - {a^2})(1 - {b^2}) + {a^2}{b^2} + 2ab\sqrt {1 - {a^2}} \sqrt {1 - {b^2}} \} $

$ - 4ab\,(\sqrt {1 - {a^2}} \sqrt {1 - {b^2}}  + ab)$

$ = \,\,2\,(1 - {b^2} - {a^2} + {a^2}{b^2}) + 2{a^2}{b^2} - 4{a^2}{b^2} - 1$

$ = \,\,2\,(1 - {a^2} - {b^2}) - 1 = 1 - 2{a^2} - 2{b^2}.$

$\Rightarrow \,\tan \,(A - B) = \tan \frac{\pi }{4}$

$ \Rightarrow \,\,\frac{{\tan A - \tan B}}{{1 + \tan A\,\tan B}} = 1$

$ \Rightarrow \,\,\tan A - \tan B - \tan A\,\tan B = 1$

$ \Rightarrow \,\,\tan A - \tan B - \tan A\,\tan B + 1 = 2$

$ \Rightarrow \,\,(1 + \,\tan A)\,\,(1 - \tan B) = 2$

$ \Rightarrow y = 2$

अत: ${(y + 1)^{y + 1}} = {(2 + 1)^{2 + 1}} = {(3)^3} = 27$.

ट्रिक : $A$ व $B$ को इस प्रकार रखें कि  $A - B = \frac{\pi }{4}$

अर्थात् $A = \frac{\pi }{4},B = 0$

$\therefore \,\,\,\left( {1 + \tan \frac{\pi }{4}} \right)\,(1 -\tan {0^o}) = 2(1) = 2$

अब $\sqrt {(4{{\sin }^4}\alpha  + {{\sin }^2}2\alpha )}  + 4{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$

$ = \sqrt {(4{{\sin }^4}\alpha  + 4{{\sin }^2}\alpha {{\cos }^2}\alpha )}  + 2.2{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$

$ = \sqrt {4{{\sin }^2}\alpha ({{\sin }^2}\alpha  + {{\cos }^2}\alpha )}  + 2\left[ {1 + \cos \left( {\frac{\pi }{2} - \alpha } \right)} \right]$

$ =  \pm 2\sin \alpha  + 2 + 2\sin \alpha $

$(-)$ चिन्ह लेने पर उत्तर $2$ और $(+)$ चिन्ह लेने पर उत्तर $2 + 4\sin \alpha $ है।

लेकिन $\pi  < \alpha  < \frac{{3\pi }}{2},$

अत: $2 - 4\sin \alpha $ उत्तर है क्योंकि यह तीसरे चतुर्थांश में ऋणात्मक होता है।

तब $\tan A = k\tan B$

या  $\frac{k}{1} = \frac{{\tan A}}{{\tan B}} = \frac{{\sin A\cos B}}{{\cos A\sin B}}$

योगन्तरानुपात से

$ \Rightarrow \frac{{k + 1}}{{k - 1}} = \frac{{\sin A\cos B + \cos A\sin B}}{{\sin A\cos B - \cos A\sin B}}$

$ = \frac{{\sin (A + B)}}{{\sin (A - B)}} = \frac{{\sin \theta }}{{\sin \phi }}$

$\Rightarrow \sin \theta  = \frac{{k + 1}}{{k - 1}}\sin \phi $.

$ \Rightarrow  {(u - \sin \theta \cos \theta )^2} = {\cos ^2}\theta ({\sin ^2}\theta  + {\sin ^2}\alpha )$

$ \Rightarrow {u^2}{\tan ^2}\theta  - 2u\tan \theta  + {u^2} - {\sin ^2}\alpha  = 0$

चूँकि $\tan \theta $ वास्तविक है।

अत: $4{u^2} - 4{u^2}({u^2} - {\sin ^2}\alpha ) \ge 0$

$ \Rightarrow {u^2} - (1 + {\sin ^2}\alpha ) \le 0$

$ \Rightarrow  |u|\, \le \sqrt {1 + {{\sin }^2}\alpha } $.

$(m + n) = a{\cos ^3}\alpha + 3a\cos \alpha \,{\sin ^2}\alpha $ $ + 3a{\cos ^2}\alpha .\sin \alpha  + a{\sin ^3}\alpha $

$ = a{(\cos \alpha + \sin \alpha )^3}$ 

एवं इसी प्रकार $(m - n) = a\,\,{(\cos \alpha - \sin \alpha )^3}$ 

अतः  ${(m + n)^{2/3}} + {(m - n)^{2/3}}$ 

$ = {a^{2/3}}{\{ \cos \alpha + \sin \alpha )^2} + {(\cos \alpha - \sin \alpha )^2}\} $ 

$ = {a^{2/3}}\{ 2({\cos ^2}\alpha + {\sin ^2}\alpha )\} = 2{a^{2/3}}$.

$ = \frac{{{{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }} + \frac{{{{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }} + \frac{{{{\tan }^2}\gamma }}{{1 + {{\tan }^2}\gamma }}$

$ = \frac{x}{{1 + x}} + \frac{y}{{1 + y}} + \frac{z}{{1 + z}}$ $(x = {\tan ^2}\alpha ,\,y = {\tan ^2}\beta ,\,z = {\tan ^2}\gamma )$

$ = \frac{{(x + y + z) + (xy + yz + zx + 2xyz) + xy + yz + zx + xyz}}{{(1 + x)(1 + y)(1 + z)}}$ 

$ = \frac{{1 + x + y + z + xy + yz + zx + xyz}}{{(1 + x)(1 + y)(1 + z)}} = 1$       $( \because xy + yz + zx + 2xyz = 1)$

या  $\sin x = 1 - {\sin ^2}x$ या $\sin x = {\cos ^2}x$

$\therefore$ ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 2$

$ = {\sin ^6}x + 3{\sin ^5}x + 3{\sin ^4}x + {\sin ^3}x - 2$

$ = {({\sin ^2}x)^3} + 3{({\sin ^2}x)^2}\sin x  + 3({\sin ^2}x){(\sin x)^2} + {(\sin x)^3} - 2$

$ = {({\sin ^2}x + \sin x)^3} - 2$

$ = {(1)^3} - 2$                     $[ \because \sin x + {\sin ^2}x = 1$ (दिया है)$]$ 

$= -1.$

$\frac{1}{{\tan \theta }} + \tan \theta  = m\,$

$\Rightarrow \,1 + {\tan ^2}\theta  = m\,\tan \theta $

$ \Rightarrow \,\,{\sec ^2}\theta  = m\,\tan \theta $…..$(i)$

व $\sec \theta  - \cos \theta  = n\,\, $

$\Rightarrow \,\,{\sec ^2}\theta  - 1 = n\,\sec \theta $

$ \Rightarrow \,\,{\tan ^2}\theta  = n\,\,\sec \theta $

$ \Rightarrow \,\,{\tan ^4}\theta  = {n^2}\,{\sec ^2}\theta  = {n^2}.\,m\,\,\tan \theta $    {$(i)$ द्वारा}

$ \Rightarrow \,\,\tan \theta  = {({n^2}m)^{1/3}}$…..$(ii)$

तथा ${\sec ^2}\theta  = m\,\,\tan \theta  = m\,{({n^2}m)^{1/3}}$     {$(i)$ व $(ii)$ से}

$\therefore$ सर्वसमिका ${\sec ^2}\theta  - {\tan ^2}\theta  = 1$ से,

$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} - {({n^2}m)^{2/3}} = 1$

$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} - n\,{(n{m^2})^{1/3}} = 1.$

$ \Rightarrow \,\,\frac{1}{x}\tan \theta - \tan \theta \,\,\cos \phi = \sin \,\phi $ 

$ \Rightarrow \,\,\frac{1}{x} = \frac{{\sin \,\phi + \cos \,\,\phi \,\tan \,\theta }}{{\tan \,\theta }}$

एवं $\tan \,\phi = \frac{{y\,\sin \,\theta }}{{1 - y\,\cos \,\theta }}$

$ \Rightarrow \tan \,\phi \, = \frac{{\sin \,\theta }}{{\frac{1}{y} - \cos \,\theta }}$

$ \Rightarrow \,\,\frac{1}{y}\tan \,\phi - \tan \,\phi \cos \theta = \sin \theta $

$ \Rightarrow \,\,\frac{1}{y}\tan \,\phi \, = \sin \,\theta + \tan \,\phi \,\cos \theta $

$\therefore \,\,\,\,\frac{1}{y} = \frac{{\sin \,\theta + \tan \,\phi \,\cos \theta }}{{\tan \,\phi }}$

अब $\frac{x}{y} = \left[ {\frac{{\tan \,\theta }}{{\sin \,\phi + \cos \,\phi \,\tan \,\theta }}} \right] \times \left[ {\frac{{\sin \,\theta + \tan \,\phi \,\cos \,\theta }}{{\tan \,\phi }}} \right]$

$ = \frac{{\tan \,\theta }}{{\tan \,\phi }}\,\left[ {\frac{{\sin \,\theta + \cos \,\theta \frac{{\sin \varphi }}{{\cos \phi }}}}{{\sin \phi + \cos \phi \frac{{\sin \theta }}{{\cos \theta }}}}} \right] $

$= \frac{{\tan \theta \,\,\cos \theta }}{{\tan \,\phi \,\cos \phi }} = \frac{{\sin \theta }}{{\sin \phi }}$

वैकल्पिक​ : $x\,\sin \,\phi = \tan \,\theta - x\,\cos \,\phi \,\tan \,\theta $

$ \Rightarrow \,x = \frac{{\tan \,\theta }}{{\sin \,\phi + \cos \,\phi \,\tan \,\theta }}$

$ = \frac{{\sin \,\theta }}{{\cos \,\theta \sin \,\phi + \cos \,\phi \,\sin \,\theta }} $

$= \frac{{\sin \,\theta }}{{\sin \,(\theta + \phi )}}$

इसी प्रकार, $y = \frac{{\sin \,\phi }}{{\sin \,(\theta + \phi )}}$;

$\therefore \,\,\frac{x}{y} = \frac{{\sin \theta }}{{\sin \phi }}.$

We have $\cos ^{4} x \leq \cos ^{2} x$

$\sin ^{2} x=\sin ^{2} x$

Adding $\sin ^{2} x+\cos ^{4} x \leq \sin ^{2} x+\cos ^{2} x$

$\therefore A \leq 1$

Again $A=t+(1-t)^{2}=t^{2}-t+1, t \geq 0$,

where minimum is $3 / 4$

Thus $3 / 4 \leq A \leq 1$.

$\sin \,(\alpha \, - \beta )\, = \frac{5}{{13}} \Rightarrow \tan \,(\alpha \, - \beta ) = \frac{5}{{12}}$

$\tan 2\alpha \, = \tan [(\alpha \, + \beta ) + (\alpha \, - \beta )]$ $\, = \,\frac{{\frac{3}{4} + \frac{5}{{12}}}}{{1 - \frac{3}{4}.\frac{5}{{12}}}} = \frac{{56}}{{33}}$

$\Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+3=0$

$\Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \beta$

$+\cos ^{2} \beta+\sin ^{2} \gamma+\cos ^{2} \alpha=0$

$\Rightarrow\left[\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma+2 \sin \gamma \sin \alpha\right]$

$+\left[\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2 \cos \alpha \cos \beta+2 \cos \beta \cos \gamma+2 \cos \gamma \cos \alpha\right]$

$=0$

$\Rightarrow[\sin \alpha+\sin \beta+\sin \gamma]^{2}+[\cos \alpha+\cos \beta+\cos \gamma]^{2}=0$

$\sin \alpha+\sin \beta+\sin \gamma=0$

$\cos \alpha+\cos \beta+\cos \gamma=0$

$\therefore \mathrm{A}$ and $\mathrm{B}$ both are true.

$ {\text { Let : } \frac{\cos ^{2} \alpha}{\cos \beta}=\cos \theta} \,\,\,\,\,\,\,\, {\frac{\sin ^{2} \alpha}{\sin \beta}=\sin \theta}$

$ {\cos ^{2} \alpha=\cos \beta \cos \theta} \,\,\,\,\,\,\,\,\, {\sin ^{2} \alpha=\sin \beta \sin \theta} $

${1=\cos (\beta-\theta)} \,\,\,\, \,\,\,\,\,\,{\Rightarrow \beta=\theta+2 n \pi} $

$\therefore $ ${\cos ^2}\alpha  = {\cos ^2}\beta \,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, {\sin ^2}\alpha  = {\sin ^2}\beta $

${\therefore \quad \frac{\cos ^{4} \beta}{\cos ^{2} \alpha}+\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}=\cos ^{2} \beta+\sin ^{2} \beta=1}$

$ = \prod\limits_{r = 1}^{59} {\frac{{\sin \left( {{{30}^\circ } + {{\rm{r}}^\circ }} \right)}}{{\cos {{\rm{r}}^\circ }}}} $

$ = \frac{{\sin {{31}^\circ } \cdot \sin {{32}^\circ } \ldots  \ldots \sin {{89}^\circ }}}{{\cos {1^\circ } \cdot \cos 2 \ldots  \ldots \cos {{59}^\circ }}} = 1$