Maths M.C.Q (1 Marks)

Put $\cos ^{-1}\left(\frac{\sqrt{5}}{3}\right)=\alpha$

$\quad \cos \alpha=\frac{\sqrt{5}}{3} \Rightarrow 0<\alpha<\frac{\pi}{2}$

$\tan \frac{\alpha}{2}=\sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}=\sqrt{\frac{1-\sqrt{5} / 3}{1+\sqrt{5} / 3}}$

$=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{(3-\sqrt{5})^{2}}{9-5}}$

$=\frac{3-\sqrt{5}}{2}$

$ \Rightarrow \,\,{\cos ^{ - 1}}(x) + {\cos ^{ - 1}}(y) + {\cos ^{ - 1}}(z) = {\cos ^{ - 1}}( - 1)$

$ \Rightarrow \,\,{\cos ^{ - 1}}(x) + {\cos ^{ - 1}}(y) = {\cos ^{ - 1}}( - 1) - {\cos ^{ - 1}}(z)$

$ \Rightarrow \,\,{\cos ^{ - 1}}(xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2})} = {\cos ^{ - 1}}\,\left\{ {( - 1)\,\,(z)} \right\}$

$ \Rightarrow \,\,xy - \sqrt {(1 - {x^2})\,\,(1 - {y^2})} = - z$

$ \Rightarrow \,\,(xy + z) = \sqrt {(1 - {x^2})\,\,(1 - {y^2})} $

Squaring both sides we get ${x^2} + {y^2} + {z^2} + 2xyz = 1$.

Trick : Put $x = y = z = \frac{1}{2},$ so that

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

Obviously $ (d)$  holds for these values of  $ x, y, z.$

==> $24{x^2} - 23x - 12 = 0 \Rightarrow (3x - 4)(8x + 3) = 0$.

${\sin ^{ - 1}}b = B,$

${\sin ^{ - 1}}c = C$

$\therefore \sin A = a,\sin B = b,\sin C = c$

and $A + B + C = \pi ,$then

$\sin 2A + \sin 2B + \sin 2C$

$ = 4\sin A\,\,\sin B\,\,\sin C$ …..$(i)$

==> $\sin A\,\cos A\, + \sin B\,\cos B + \sin \,C\,\cos C$

$= 2\sin A\sin B\sin C$

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

$ = 2\sin A\sin B\sin C.$ ……$(ii)$

==> $a\sqrt {(1 - {a^2})} + b\sqrt {(1 - {b^2})} + c\sqrt {{{(1 - c)}^2}} = 2abc$,

while ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi $.

and $\gamma = {\cos ^{ - 1}}\sqrt {1 - q} \,\,{\rm{or}}\cos \alpha = \sqrt p ;\,\,\cos \beta = \sqrt {1 - p} $

and $\cos \gamma = \sqrt {1 - q.} $

Therefore $\sin \alpha = \sqrt {1 - p} ,$ $\sin \beta = \sqrt p $ and $\sin \gamma = \sqrt q $.

The given equation may be written as $\alpha + \beta + \gamma = \frac{{3\pi }}{4}$ 

or $\alpha + \beta = \frac{{3\pi }}{4} - \gamma $ or $\cos (\alpha + \beta ) = \cos \left( {\frac{{3\pi }}{4} - \gamma } \right)$

==> $\cos \alpha \,\,\,\cos \beta - \sin \alpha \sin \beta = $

$\cos \left\{ {\pi - \left( {\frac{\pi }{4} + \gamma } \right)} \right\} = - \cos \left( {\frac{\pi }{4} + \gamma } \right)$

==> $\sqrt {p\,} \sqrt {1 - p} - \sqrt {1 - p} \sqrt p $

$ = - \left( {\frac{1}{{\sqrt 2 }}\sqrt {1 - q} - \frac{1}{{\sqrt 2 }}.\sqrt q } \right)$

==> $0 = \sqrt {1 - q} - \sqrt q $

==> $1 - q = q$ ==> $q = \frac{1}{2}.$

$ = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)} \right]$

(Let $a = \tan \theta $)

$ = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}(\sin 2\theta ) + \frac{1}{2}{{\cos }^{ - 1}}(\cos 2\theta )} \right]$

$ = \tan (2\theta ) = \tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{2a}}{{1 - {a^2}}}$

Trick : Put $a = 0$, then tan $(0+0)=0;$ which is given by $(a)$ and $(c)$.

Again put $a = 1$, then $\tan \left( {\frac{\pi }{4} + \frac{\pi }{4}} \right) = \infty $, which is given by $(c).$

Let $\frac{1}{2}{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3} = \theta $

$\Rightarrow \cos 2\theta = \frac{{\sqrt 5 }}{3}$

But $\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} $

$\Rightarrow \frac{{\sqrt 5 }}{3} = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

==>$\sqrt 5 + \sqrt 5 {\tan ^2}\theta = 3 - 3{\tan ^2}\theta $

==> $(\sqrt 5 + 3){\tan ^2}\theta = 3 - \sqrt 5 $

$\Rightarrow {\tan ^2}\theta = \frac{{3 - \sqrt 5 }}{{3 + \sqrt 5 }}$

==> ${\tan ^2}\theta = \frac{{{{(3 - \sqrt 5 )}^2}}}{4} $

$\Rightarrow \tan \theta = \frac{{3 - \sqrt 5 }}{2}$

On rationalising

==> $\tan \theta = \frac{{3 - \sqrt 5 }}{2} \times \frac{{3 + \sqrt 5 }}{{3 + \sqrt 5 }} = \frac{2}{{3 + \sqrt 5 }}$.

$ = {\cot ^{ - 1}}\left[ {\frac{{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x} )}}{{(\sqrt {1 - \sin x} - \sqrt {1 + \sin x} )}}.\frac{{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x} )}}{{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x} )}}} \right]$

$= {\cot ^{ - 1}}\left[ {\frac{{(1 - \sin x) + (1 + \sin x) + 2\sqrt {1 - {{\sin }^2}x} }}{{(1 - \sin x) - (1 + \sin x)}}} \right]$

$= {\cot ^{ - 1}}\left[ {\frac{{2(1 + \cos x)}}{{ - 2\sin x}}} \right] = {\cot ^{ - 1}}\left[ { - \frac{{2{{\cos }^2}(x/2)}}{{2\sin (x/2)\cos (x/2)}}} \right]$

$= {\cot ^{ - 1}}\left( { - \cot \frac{x}{2}} \right) = {\cot ^{ - 1}}\left[ {\cot \left( {\pi - \frac{x}{2}} \right)} \right] = \pi - \frac{x}{2}$.

Trick : Put $x = \frac{\pi }{4}$, so that the expression becomes

${\cot ^{ - 1}}\left[ {\frac{{\sqrt {\sqrt 2 - 1} + \sqrt {\sqrt 2 + 1} }}{{\sqrt {\sqrt 2 - 1} - \sqrt {\sqrt 2 + 1} }}} \right]$

$= {\cot ^{ - 1}}\left[ {\frac{{\sqrt 2 - 1 + \sqrt 2 + 1 + 2\sqrt {2 - 1} }}{{\sqrt 2 - 1 - \sqrt 2 - 1}}} \right]$

$= {\cot ^{ - 1}}\left[ {\frac{{2\sqrt 2 + 2}}{{ - 2}}} \right] = {\cot ^{ - 1}}( - 1 - \sqrt 2 ) = 157.5^\circ $.

$p = q = r = \frac{1}{2}$.

Then, ${p^2} + {q^2} + {r^2} + 2pqr$

$= {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^2} + 2.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}$

$= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{2}{8} =1.$

Let $\frac{1}{2}{\cos ^{ - 1}}\frac{a}{b} = \theta $

$\Rightarrow \cos 2\theta = \frac{a}{b}$

Thus, $\tan \left[ {\frac{\pi }{4} + \theta } \right] + \tan \left[ {\frac{\pi }{4} - \theta } \right]$

$=  \frac{{1 + \tan \theta }}{{1 - \tan \theta }} + \frac{{1 - \tan \theta }}{{1 + \tan \theta }} = \frac{{{{(1 + \tan \theta )}^2} + {{(1 - \tan \theta )}^2}}}{{(1 - {{\tan }^2}\theta )}}$

$=  \frac{{1 + {{\tan }^2}\theta + 2\tan \theta + 1 + {{\tan }^2}\theta - 2\tan \theta }}{{(1 + {{\tan }^2}\theta )}}$

$=  \frac{{2(1 + {{\tan }^2}\theta )}}{{1 - {{\tan }^2}\theta }} = 2\sec 2\theta = \frac{2}{{\cos 2\theta }}$

$=  \frac{2}{{a/b}} = \frac{{2b}}{a}$.

We have,

$\sin ^{-1}(\sin 1 \cos 4+\cos 1 \sin 4)$

$\quad=\sin ^{-1}(\sin 5)$

$[\because \sin A \cos B+\cos A \sin B=\sin (A+B)]$

$=5-2 \pi=5-2(3.14)=-1.28$

i.e., nearest integer is $-1$.

$ \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{n}=\frac{\pi}{4} $

$ \tan ^{-1}\left(\frac{46}{48}\right)+\tan ^{-1} \frac{1}{n}=\frac{\pi}{4} $

$ \tan ^{-1}\left(\frac{23}{24}\right)+\tan ^{-1} \frac{1}{n}=\frac{\pi}{4} $

$ \tan ^{-1} \frac{1}{n}=\tan ^{-1} 1-\tan ^{-1} \frac{23}{24} $

$ \tan ^{-1} \frac{1}{n}=\tan ^{-1}\left(\frac{1-\frac{23}{24}}{1+\frac{23}{24}}\right) $

$ \tan ^{-1} \frac{1}{n}=\tan ^{-1}\left(\frac{1}{\frac{24}{47}}\right. $

$ \tan ^{-1} \frac{1}{n}=\tan ^{-1} \frac{1}{47} $

$ n=47$

$ \Rightarrow \quad \pi+\cos ^{-1} x=\frac{2 \pi}{5}$

$ \Rightarrow \quad \cos ^{-1} x=\frac{-3 \pi}{5}$

Not possible

Ans. $0$

$\text { and } b=\cos ^{-1}(\cos 5)=2 \pi-5 $

$\therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 $

$=8 \pi^2-40 \pi+50$

$ \cos (2 \theta)=\frac{1}{9} $

$ \sin \theta= \pm \frac{2}{3}$

as $\mathrm{m}$ and $\mathrm{n}$ are co-prime natural numbers,

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

i.e. $m=2, n=3$

So, the quadratic equation becomes $2 x^2-3 x+1=$ 0 whose roots are $\alpha=1, \beta=\frac{1}{2}$ $\left(1, \frac{1}{2}\right)$ lies on $5 x+8 y=9$

$\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$

$(\alpha+\beta)^2-\gamma^2=3 \alpha \beta$

$\alpha^2+\beta^2-\gamma^2=\alpha \beta$

$\frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2}$

$\Rightarrow \cos \mathrm{C}=\frac{1}{2}$

$\operatorname{sinC}=\gamma$

$\cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2}$

$\gamma=\frac{\sqrt{3}}{2}$

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

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

$\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\frac{(1+x)\left(\sqrt{x^2+x+1}\right)}{\left(x^2+x\right)(\sqrt{x})}$

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

$ \tan (A+B)=\frac{\sqrt{x^2+x+1}}{x \sqrt{x}}=\tan C $

$ A+B=C$

$\Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x$

Taking tan both sides

$\Rightarrow 2 x=\frac{1-x}{1+x} $

$\Rightarrow 2 x^2+3 x-1=0$

$x=\frac{-3 \pm \sqrt{9+8}}{8}=\frac{-3 \pm \sqrt{17}}{8} $

$ \text { Only possible } x=\frac{-3+\sqrt{17}}{8}$

$ \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 $

$ -4 \leq 2-|x| \leq 4 $

$ -6 \leq-|x| \leq 2 $

$ -2 \leq|x| \leq 6 $

$ |x| \leq 6 $

$ \Rightarrow x \in[-6,6]$        $................(1)$

Now, $3-x \neq 1$

And $x \neq 2$        $................(2)$

and $3-x>0$

$x<3$           $............(3)$

From $(1)$, $(2)$ and $(3)$

$ \Rightarrow x \in[-6,3)-\{2\} $

$ \alpha=6$

$ \beta=3 $

$ \gamma=2 $

$ \alpha+\beta+\gamma=11$

$ \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha $

$ \alpha \in\left[-\frac{\pi}{2}, \pi\right], \frac{\pi}{2}+\alpha \in\left[0, \frac{3 \pi}{2}\right] $

$ \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha $

$ x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha $

$ (x y+\sin \alpha)=\left(1-x^2\right)\left(1-y^2\right) $

$ x^2 y^2+2 x y \operatorname{sina}+\sin ^2 a=1-x^2-y^2+x^2 y^2 $

$ x^2+y^2+2 x y \sin \alpha=1-\sin ^2 \alpha $

$ x^2+y^2+2 x y \sin \alpha=\cos ^2 \alpha$

Min. value of $\cos ^2 \alpha=0$

At $\alpha=\frac{\pi}{2}$

Option ($2$) is correct

$=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{3}$

$\therefore \sin ^{-1} \frac{\alpha}{17}=\tan ^{-1} \frac{77}{36}-\tan ^{-1} \frac{3}{4}=\tan ^{-1}\left(\frac{\frac{77}{36}-\frac{3}{4}}{1+\frac{77}{36} \cdot \frac{3}{4}}\right)$

$\sin ^{-1} \frac{\alpha}{17}=\tan ^{-1} \frac{8}{15}=\sin ^{-1} \frac{8}{17}$

$\Rightarrow \frac{\alpha}{17}=\frac{8}{17} \Rightarrow \alpha=8$

$\therefore \sin ^{-1}(\sin 8)+\cos ^{-1}(\cos 8)$

$=3 \pi-8+8-2 \pi$

$=\pi$

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

$x=2-\sqrt{3}$

Case $II:$ $x < 0$

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

$x=\frac{-1}{\sqrt{3}} \Rightarrow \alpha=2$

$\therefore \tan ^{-1}\left(\frac{ a _2- a _1}{1+ a _1 a _2}\right)+\tan ^{-1}\left(\frac{ a _3- a _2}{1+ a _2 a _3}\right)+\ldots . .+\tan ^{-1}\left(\frac{ a _{2022}- a _{2021}}{1+ a _{2021} a _{2022}}\right)$

$=\left[\left(\tan ^{-1} a _2\right)-\tan ^{-1} a _1\right]+\left[\tan ^{-1} a _3-\tan ^{-1} a _2\right]+\ldots . .$

$+\left[\tan ^{-1} a _{2022}-\tan ^{-1} a _{2021}\right]$

$=\tan ^{-1} a _{2022}-\tan ^{-1} a _1$

$=\tan ^{-1}(2022)-\tan ^{-1} 1=\tan ^{-1} 2022-\frac{\pi}{4} \text { (option 3) }$

$=\left(\frac{\pi}{2}-\cot ^{-1}(2022)\right)-\frac{\pi}{4}$

$=\frac{\pi}{4}-\cot ^{-1}(2022) \text { (option 1) }$

$\Rightarrow \sin ^{-1} \sin \theta > \frac{\pi}{4}$

$\Rightarrow \sin \theta > \frac{1}{\sqrt{2}}$

$\text { So, } \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$

$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=( a , b )$

$b - a =\frac{\pi}{2}=\alpha-\beta$

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

$\Rightarrow \alpha x^2+\beta x+\sin ^{-1}\left[(x-3)^2+1\right]+\cos ^{-1}\left[(x-3)^2+1\right]=0$

$x =3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$

$\Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0$

$\Rightarrow 12 \alpha-\pi=0$

$\alpha=\frac{\pi}{12}$

$\cos ^{-1}(2 x)-\cos ^{-1}\left(2\left(1-x^2\right)-1\right)=\pi$

$\cos ^{-1}(2 x)-\cos ^{-1}\left(1-2 x^2\right)=\pi$

$-\cos ^{-1}\left(1-2 x^2\right)=\pi-\cos ^{-1}(2 x)$

Taking $\cos$ both sides we get

$\operatorname{Cos}\left(-\cos ^{-1}\left(1-2 x^2\right)\right)=\cos \left(\pi-\cos ^{-1}(2 x)\right)$

$1-2 x^2=-2 x$

$2 x^2-2 x-1=0$

On solving, $x=\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}$

As $x=[-1 / 2,1 / 2], x=\frac{1+\sqrt{3}}{2}=$ rejected

So $x=\frac{1-\sqrt{3}}{2} \Rightarrow x^2-1=-\sqrt{3} / 2$

$=2 \sin ^{-1}\left(x^2-1\right)=2 \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-2 \pi}{3}$

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

$\tan ^{-1} x =\theta \in\left(0, \frac{\pi}{4}\right) \therefore x =\tan \theta$

$2 \tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)=\cos ^{-1}(\cos 2 \theta)$

$2\left(\frac{\pi}{4}-\theta\right)=2 \theta \therefore 4 \theta=\frac{\pi}{2} \therefore \theta=\frac{\pi}{8}$

$x =\tan \frac{\pi}{8} \therefore x =\sqrt{2}-1 \simeq 0.414$

$\because \frac{t}{\sqrt{t^2+1}} \in(-1,1)$

$\quad \sin ^{-1}\left(\frac{(x+1)}{\sqrt{(x+1)^2+1}}\right)=\sin ^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)+\frac{\pi}{4}$

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

$=\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{ x ^2+1}}+\frac{ x }{\sqrt{ x ^2+1}}\right)$

$\frac{( x +1)}{\sqrt{( x +1)^2+1}}=\frac{1}{\sqrt{2}}\left(\frac{1+ x }{\sqrt{ x ^2+1}}\right)$

After solving this equation, we get

$x=-1 \text { or } x=0$

$S=\{-1,0\}$

$\sum_{x \in R }\left(\sin \left(\left(x^2+ x +5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+ x +5\right) \pi\right)\right)$

$=\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]+\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]$

$=4$

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

$E =\cos ^{-1}\left(\frac{3}{10} \cos \theta+\frac{2}{5} \sin \theta\right)$

$=\cos ^{-1}\left(\frac{3}{10} \times \frac{3}{5}+\frac{2}{5} \cdot \frac{4}{5}\right)$

$=\cos ^{-1}\left(\frac{9}{50}+\frac{8}{25}\right)=\cos ^{-1}\left(\frac{25}{50}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

$\sin ^{-1} \sin \left(\frac{2 \pi}{3}\right)=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}$

$\cos ^{-1}\left(\cos \frac{2 \pi}{6}\right)=2 \pi-\frac{7 \pi}{6}=\frac{5 \pi}{6}$

$\tan ^{-1} \tan \left(\frac{3 \pi}{4}\right)=\frac{3 \pi}{4}-\pi=\frac{-\pi}{4}$

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1} \cos \frac{7 \pi}{6}+\tan ^{-1} \tan \frac{3 \pi}{4}$

$=\frac{11 \pi}{12}$

$\left(x^{2}-4 x+2\right)^{2} \leq\left(x^{2}+3\right)^{2}$

$\left(x^{2}-4 x+2\right)^{2}-\left(x^{2}+3\right)^{2} \leq 0$

$\left(2 x^{2}-4 x+5\right)(-4 x-1) \leq 0$

$-4 x-1 \leq 0 \rightarrow x \geq-\frac{1}{4}$

$=-2\left[\frac{1}{\sqrt{1-x^{2}}}+\frac{2}{1+x^{2}}+3 x+1\right]$

$f^{\prime}(x)<0 \Rightarrow f(x)$ is a dec. function

$f(1)=\pi+5$

$f(-1)=5 \pi+5$

Range : $[ a , b ] \equiv[\pi+5,5 \pi+5]$

$a =\pi+5, b =5 \pi+5 \Rightarrow 4 a - b =11-\pi \text {. }$

$\tan ^{-1}\left(\frac{1-\sqrt{2}}{1}\right)=-\frac{\pi}{8}$

$=\tan \left[2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right]$

$=2$

$=50 \tan \left(\tan ^{-1} \frac{1}{2}+2\left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} 2\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$

$\left.=50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \cdot \frac{\pi}{2}\right)\right)+4 \sqrt{2} \times \frac{1}{\sqrt{2}}$

$=50\left(\tan \tan ^{-1} \frac{1}{2}\right)+4$

$=25+4=29$

$\left(x^{2}+1\right) * 1=x *(2)$

$\left(x^{2}+1\right)^{2}+1=x^{2}+8$

$x^{4}+x^{2}-6=0 \Rightarrow\left(x^{2}+3\right)\left(x^{2}-2\right)=0$

$x^{2}=2$

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

$=\frac{\pi}{3}$

$\left(\tan ^{-1} x+\cot ^{-1} x\right)-3 \tan ^{-1} x \cdot \cot ^{-1} x\left(\tan ^{-1} x+\cot ^{-1} x\right)$

$=\frac{\pi^{3}}{8}-\frac{3 \pi}{2} \tan ^{-1} x\left(\frac{\pi}{2}-\tan ^{-1} x\right)$

$=\frac{3 \pi}{2}\left(\tan ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{3}}{32}$

$\Rightarrow \frac{\pi^{3}}{32} \leq S <\frac{7}{8} \pi^{3}$

$=\frac{\pi^{3}}{32} \leq K \pi^{3}<\frac{7}{8} \pi^{3}$

$\frac{1}{32} \leq K <\frac{7}{8}$

$=\tan ^{-1}(n+1)-\tan ^{-1} n$

so, $\sum\limits_{n=1}^{50}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)$

$=\tan ^{-1} 51-\tan ^{-1} 1$

$\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)=\cot \left(\tan ^{-1} 51+\tan ^{-1} 1\right)$

$=\frac{1}{\tan \left(\tan ^{-1} 51-\tan ^{-1} 1\right)}=\frac{1+51 \times 1}{51-1}=\frac{52}{50}=\frac{26}{25}$

$=\tan ^{-1}( r +2)-\tan ^{-1}( r +1)$

$T _{1}=\tan ^{-1} 3-\tan ^{-1} 2$

$T _{2}=\tan ^{-1} 4-\tan ^{-1} 3$

$T _{ n }=\tan ^{-1}( n +2)-\tan ^{-1}( n +1)$

_______________________________

$S _{ n }=\tan ^{-1}( n +2)-\tan ^{-1} 2=\tan ^{-1}\left(\frac{ n +2-2}{1+2( n +2)}\right)$

$=\tan ^{-1}\left(\frac{n}{2 n+5}\right)$

$\lim\limits _{ n \rightarrow \infty} 6 \tan ^{}\left(\tan ^{-1}\left(\frac{ n }{2 n +5}\right)\right)$

$=\lim \limits_{n \rightarrow \infty} \frac{6 n}{2 n+5}=\frac{6}{2}=3$

$\because \quad x=\sin \left(2 \tan ^{-1} \alpha\right)=\frac{2 \alpha}{1+\alpha^{2}}$

and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)=\sin \left(\sin ^{-1} \frac{1}{\sqrt{5}}\right)=\frac{1}{\sqrt{5}}$

Now, $y^{2}=1-x$

$\frac{1}{5}=1-\frac{2 \alpha}{1+\alpha^{2}}$

$1+\alpha^{2}=5+5 \alpha^{2}-10 \alpha$

$2 \alpha^{2}-5 \alpha+2=0$

$\therefore \quad \alpha=2, \frac{1}{2}$

$\therefore \quad \sum_{\alpha \in S} 16 \alpha^{3}=16 \times 2^{3}+16 \times \frac{1}{2^{3}}$

$\sin ^{-1} x=k \alpha$

$\cos ^{-1} x=k \beta$

$k=\frac{\pi}{2(\alpha+\beta)}$

$\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)=\sin \left(4 \sin ^{-1} x\right)$

$=2 \sin \left(2 \sin ^{-1} x\right) \cos \left(2 \sin ^{-1} x\right)$

$=4 x \sqrt{1-x^{2}}\left(1-2 x^{2}\right)$

$P_{1}:-1 \leq\left[2 x^{2}-3\right]<1$

$\Rightarrow-1 \leq 2 x ^{2}-3<2$

$\Rightarrow 2<2 x ^{2}<5$

$\Rightarrow 1< x ^{2}<\frac{5}{2}$

$\Rightarrow P _{1}: x \in\left(-\sqrt{\frac{5}{2}},-1\right) \cup\left(1, \sqrt{\frac{5}{2}}\right)$

$P_{2}: x^{2}-5 x+5>0$

$\Rightarrow\left( x -\left(\frac{5-\sqrt{5}}{2}\right)\right)\left( x -\left(\frac{5+\sqrt{5}}{2}\right)\right)>0$

$P_{3}: \log _{\frac{1}{2}}\left(x^{2}-5 x+5\right)>0$

$x ^{2}-5 x -5<1$

$x ^{2}-5 x +4<0$

$P _{3}: x \in(1,4)$

So, $P _{1} \cap P _{2} \cap P _{3}=\left(1, \frac{5-\sqrt{5}}{2}\right)$

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

$\cos ^{-1} x-\pi+2 \cos ^{-1} x=\cos ^{-1} 2 x$

$3 \cos ^{-1} x=\pi+\cos ^{-1} 2 x$

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

$4 x^{3}-3 x=-2 x$

$4 x^{3}=x \Rightarrow x=0, \pm \frac{1}{2}$

All satisfy the original equation

$\text { sum }=-\frac{1}{2} \text { to }+\frac{1}{2}=0$

$\frac{x^{2}-3 x+2}{x^{2}+2 x+7} \geq-1$ and $\frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1$

$2 x^{2}-x+9 \geq 0$ and $5 x \geq-5 \Rightarrow x \geq-1$ $x \in R$

Hence Domain $x \in[-1, \infty)$

$\cot \left(\tan ^{-1} \sqrt{1-x^{2}}\right)=\cot ^{-1} \cot ^{-1}\left(\sqrt{\left.\frac{1}{\sqrt{1-x^{2}}}\right)=\frac{1}{\sqrt{1-x^{2}}}}\right.$

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

$\frac{\sqrt{1-2 x^{2}}}{\sqrt{1-x^{2}}}=k$

$1-2 x^{2}=k^{2}\left(1-x^{2}\right)$

$\left(k^{2}-2\right) x^{2}=k^{2}-1$

$\alpha=\sqrt{\frac{k^{2}-1}{k^{2}-2}} \Rightarrow \alpha^{2}=\frac{k^{2}-1}{k^{2}-2}$

$\beta=\sqrt{\frac{k^{2}-1}{k^{2}-2}} \Rightarrow \beta^{2}=\frac{k^{2}-1}{k^{2}-2}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=2\left(\frac{k^{2}-2}{k^{2}-1}\right) \& \frac{\alpha}{\beta}=-1$

Sum of roots $=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{\alpha}{\beta}=b$

$\frac{2\left(k^{2}-2\right)}{k^{2}-1}-1=b \ldots \ldots(1)$

Product of roots $=\left(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\right) \frac{\alpha}{\beta}=-5$

$\frac{2\left(k^{2}-2\right)}{k^{2}-1}(-1)=-5$

$2 k^{2}-4=5 k^{2}-5$

$3 k^{2}=1 \Rightarrow k^{2}=\frac{1}{3} \ldots$ Put in $(1)$

$b=\frac{2\left(k^{2}-2\right)}{k^{2}-1}-1=5-1=4$

$\frac{b}{k^{2}}=\frac{4}{\frac{1}{3}}=12$

$\operatorname{cosec}\left[\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{3}{4}\right)\right]$

$=\operatorname{cosec}\left[\tan ^{-1}\left(\frac{56}{33}\right)\right]=\frac{65}{56}$

$\tan ^{-1} \frac{\frac{6}{5}}{1-\frac{9}{25}}=\underbrace{\tan ^{-1} \frac{15}{8}+\tan ^{-1} \frac{5}{12}}_{x\,>\,0, y\,>\,0, x y\,<\,1}$

$\tan ^{-1} \frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8} \cdot \frac{5}{12}}=\tan ^{-1} \frac{220}{21}$

$\tan \left(\tan ^{-1} \frac{220}{21}\right)=\frac{220}{21}$

$\sin 4 \theta=\frac{\sqrt{63}}{8}$

$\cos 4 \theta=\frac{1}{8}$

$2 \cos ^{2} 2 \theta-1=\frac{1}{8}$

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

$\cos 2 \theta=\frac{3}{4}$

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

$\cos ^{2} \theta=\frac{7}{8}$

$\cos \theta=\frac{\sqrt{7}}{2 \sqrt{2}}$

$\tan \theta=\frac{1}{\sqrt{7}}$

$\Rightarrow(2 \pi-5)+(6-2 \pi)-(12-4 \pi)$

$\Rightarrow 4 \pi-11$