Maths M.C.Q (1 Marks)

But $x$ is real, therefore ${B^2} - 4AC \ge 0$

$ \Rightarrow {\cos ^2}\theta \ge 4(1)(1) \Rightarrow {\cos ^2}\theta \ge 4$, which is impossible.

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

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

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

$\Rightarrow a = b.$

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

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

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

which is true, if $t = \tan \theta = \frac{4}{3}$ or $\tan \theta = \frac{{2m}}{{{m^2} - 1}}$.

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

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

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

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

From $(i)$ and $(ii),$

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

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

$ \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 $

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

$ = \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)}}$

$= \tan \frac{A}{2}$. 

Trick : Put $A = {60^o}.$ 

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

which is given by option $(c)$,

$i.e.$ $\tan \frac{{{{60}^o}}}{2} = \frac{1}{{\sqrt 3 }}$

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

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

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

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

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

Solving $(i)$ and $(iii)$, 

we get $y = \sin \theta $ and $x = 2\,\,\cos \theta $

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

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

$ = \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)$

Adding $(i)$ and $(ii)$ we get, $xy + 1 + (x - y) = 0$ 

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

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

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

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

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

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

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

$ = \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)$.

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

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

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

$\therefore$ ${\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$.

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

Now from $(ii)$, $x\,\sin \,\alpha = y\,\cos \,\alpha $

Putting in $(i),$ we get

$ \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 $ and $x = \cos \,\alpha $

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

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

Now ${({b^2} - 1)^2}({a^2} + 4)$

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

$ = {[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$. 

Trick : Obviously the value of expression ${({b^2} - 1)^2}({a^2} + 4)$ is independent of $\theta $,

therefore put any suitable value of $\theta $.

Let $\theta = 45^\circ $, we get $a = 0,\;b = \sqrt 2 $

so that ${[{(\sqrt 2 )^2} - 1]^2}$ $({0^2} + 4) = 4.$

and $\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}$.

$ = 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$.

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

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

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

$\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 }}$

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

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

We know that $\sin {90^o} = 1$ or ${\sin ^2}{90^o} = 1$. 

Similarly, $\sin {45^o} = \frac{1}{{\sqrt 2 }}{\rm{or}}\,{\rm{si}}{{\rm{n}}^{\rm{2}}}{45^o} = \frac{1}{2}$ 

and the angles are in $A.P. $ of $18$ terms. 

We also know that ${\sin ^2}{85^o} = {[\sin ({90^o} - {5^o})]^2}$$ = {\cos ^2}{5^o}.$ 

Therefore from the complementary rule, we find 

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

Therefore, 

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

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

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

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

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

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

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

from (i) and (ii),

$ \Rightarrow B = \frac{{19\pi }}{{24}}$.

$\therefore \quad \sin \theta_{1}=\sin \theta_{2}=\sin \theta_{3}=1$

$\Rightarrow \quad \theta_{1}=\theta_{2}=\theta_{3}=\frac{\pi}{2}$

$ \therefore \quad \cos \theta_{1}=\cos \theta_{2}=\cos \theta_{3}=0 $

$ \therefore \quad \cos \theta_{1} =\cos \theta_{2}=\cos \theta_{3}=0 $

$=\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\tan 40^{\circ} \tan 20^{\circ}} $

$ \therefore  \sqrt{3}-\sqrt{3} \tan 40^{\circ} \tan 20^{\circ}=\tan 40^{\circ}+\tan 20^{\circ} $

Hence $\tan {40^\circ } + \tan {20^\circ } + \sqrt 3 \tan {40^\circ }\tan {20^\circ }$

$=\sqrt{3}$

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

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

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

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

$\Rightarrow \,x = - y$ or $x + y = 2\alpha $

Clearly, $x = - y$ satisfies $(i); $

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

which is zero, if, $\sin \left( {\frac{\pi }{4} + x - y} \right) = 0$

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

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

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

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

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

$ \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}$

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

$ = \frac{{2\,\frac{C}{D}}}{{1 + \frac{{{C^2}}}{{{D^2}}}}} = \frac{{2CD}}{{({C^2} + {D^2})}}$.

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

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

$ \Rightarrow \,\,\sin \,(A - B) = 0$

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

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

$ \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} $

$= \sqrt 3 .$

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

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

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

$ = \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)$

$= - \frac{{16}}{{65}}$

(Since $A$ lies in first quadrant અને $B$ lies in third quadrant).

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

$ \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$.

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

The value of the given expression

$ = \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)$.

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

$ = \,\,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 \cos \alpha = \frac{8}{{17}},\sin \beta = \frac{{12}}{{13}}$

and $\cos \beta = - \frac{5}{{13}}$

==> $\pi < \beta < \frac{{3\pi }}{2}$, 

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

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

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

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

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

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

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

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

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

$ \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)$

Hence, $\cot \,\left( {\frac{{A + B}}{2}} \right) = \frac{{m + 1}}{{m - 1}}\tan \frac{{B - A}}{2}$.

$ = \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}}$

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

$ = \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}}$.

$ = 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}\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 }} $

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

$ = \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}$.

$ = {\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{{\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$.

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

Squaring, ${\sin ^2}\theta + {\sin ^2}\phi + 2\sin \theta \sin \phi = {a^2}$

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

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

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

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

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

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

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

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

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

Trick : Put $\theta = \frac{\pi }{2},\phi = {0^o}$, then $a = 1 = b$

$\tan \frac{{\theta - \phi }}{2} = 1$, which is given by $(a)$ and $(b).$

Again putting $\theta = \frac{\pi }{4} = \phi $,

we get $\tan \frac{{\theta - \phi }}{2} = 0$, which is given by $(b).$

${=\frac{(1-\sin \alpha)-(1+\sin \alpha)}{\sqrt{1-\sin ^{2} \alpha}}} $

${=\frac{-2 \sin \alpha}{|\cos \alpha|}=\frac{-2 \sin \alpha}{-\cos \alpha}} $

$  \left[\because {\frac{\pi }{2} < \alpha  < \pi \therefore \cos \alpha {\rm{ is  - ve }}} \right] = 2{\rm{tan}}\alpha $