Answer:
Option B
Explanation:
Here, innermost function is inverse
$\therefore$ put, $\tan^{-1}y=\theta\Rightarrow\tan\theta=y$
$\Rightarrow \left[\frac{1}{y^{2}}\left(\frac{\cos(\tan^{-1}y) +y \sin(\tan ^{-1}y)}{\cot^{-1}(\tan^{-1}y) +\tan(\sin ^{-1}y)}\right)^{2}+y^{4}\right]^{1/2}$
$\Rightarrow \left[\frac{1}{y^{2}}\left(\frac{\frac{1}{\sqrt{1+y^{2}}}+\frac{y^{2}}{\sqrt{1+y^{2}}}}{\frac{\sqrt{1-y^{2}}}{y}+\frac{y}{\sqrt{1-y^{2}}}}\right)^{2}+y^{4}\right]^{1/2}$
$\Rightarrow\left[\frac{1}{y^{2}}.y^{2}(1-y^{4})+y^{4}\right]^{1/2}=1$
cos x +cos y= -cos z
sinx +sin y =-sinz
on squaring and adding , we get
(Q) $\cos^{2}x+\sin^{2}x+\cos^{2}y+\sin^{2}y+2\cos x\cos y+2\sin x \sin y=1$ $\Rightarrow$ $ 2+2(\cos (x-y)=1$$\Rightarrow$ $(\cos (x-y)=-\frac{1}{2}$ $\Rightarrow$ $2\cos^{2}(\frac{x-y}{2})-1=-\frac{1}{2}$$\Rightarrow$ $2\cos^{2}(\frac{x-y}{2})=\frac{1}{2}$ $\Rightarrow\cos^{}(\frac{x-y}{2})=\frac{1}{2}$ (R) $\cos 2x\left( \cos(\frac{\pi}{4}-x)-\cos(\frac{\pi}{4}+x)\right)+2\sin^{2}x=2\sin x\cos x$$\cos 2x\left( \sqrt{2}\sin x\right)+2\sin^{2}x=2\sin x\cos x$$\sqrt{2}\sin x\left( \cos 2 x+\sqrt{2}\sin^{}x-\sqrt{2}\cos x\right)=0$ $\Rightarrow $ $\sin x=0, (\cos x-\sin x)(\cos x+\sin x-\sqrt{2})=0$ $\Rightarrow$ $\sec x=1 or \frac{1}{\sqrt{2}}$ (S) $\cot (\sin^{-1}\sqrt{1-x^{2}})=\sin(\tan ^{-1}(x\sqrt{6}))$ $\Rightarrow $ $\frac{x}{\sqrt{1-x^{2}}}=\frac{x\sqrt{6}}{\sqrt{1+6x^{2}}}$
= 12 x2 =5 $x= \sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{2\sqrt{3}}$
P:4,Q:3,R:2,S:1