Match List I with List II and select the corr | vedclass

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(D) $(P)$ $\left(\frac{1}{y^2}\left(\frac{\cos (	an ^{-1} y)+y \sin (	an ^{-1} y)}{\cot (\sin ^{-1} y)+	an (\sin ^{-1} y)}ight)^2+y^4ight)^{1 /

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$3 \quad 4 \quad 1 \quad 2$

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(D) $(P)$ $\left(\frac{1}{y^2}\left(\frac{\cos (	an ^{-1} y)+y \sin (	an ^{-1} y)}{\cot (\sin ^{-1} y)+	an (\sin ^{-1} y)}ight)^2+y^4ight)^{1 / 2}$$= \left[\frac{1}{y^2}\left[\frac{\left(\frac{1}{\sqrt{1+y^2}}+\frac{y \cdot y}{\sqrt{1+y^2}}ight)}{\left(\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}ight)}ight]^2+y^4ight]^{1 / 2}$$= \left(\frac{1}{y^2} \cdot y^2(1-y^4)+y^4ight)^{1 / 2} = 1$$(Q)$ $\cos x+\cos y=-\cos z$ and $\sin x+\sin y=-\sin z$.Squaring and adding: $(\cos x+\cos y)^2 + (\sin x+\sin y)^2 = \cos^2 z + \sin^2 z = 1$.$2 + 2\cos(x-y) = 1 \Rightarrow \cos(x-y) = -1/2$.Using $\cos(x-y) = 2\cos^2(\frac{x-y}{2}) - 1 = -1/2 \Rightarrow 2\cos^2(\frac{x-y}{2}) = 1/2 \Rightarrow \cos^2(\frac{x-y}{2}) = 1/4 \Rightarrow \cos(\frac{x-y}{2}) = 1/2$.$(R)$ $\cos 2 x(\cos (\frac{\pi}{4}-x)-\cos (\frac{\pi}{4}+x)) = 2 \sin x \cos x - 2 \sin^2 x = 2 \sin x (\cos x - \sin x)$.$\cos 2 x(\sqrt{2} \sin x) = 2 \sin x (\cos x - \sin x)$.$\sqrt{2} \sin x [\cos 2 x - \sqrt{2}(\cos x - \sin x)] = 0$.$\sin x = 0$ or $\cos^2 x - \sin^2 x = \sqrt{2}(\cos x - \sin x) \Rightarrow (\cos x - \sin x)(\cos x + \sin x - \sqrt{2}) = 0$.$\sec x = \sqrt{2}$.$(S)$ $\cot (\sin ^{-1} \sqrt{1-x^2}) = \frac{x}{\sqrt{1-x^2}}$.$\sin (	an ^{-1}(x \sqrt{6})) = \frac{x \sqrt{6}}{\sqrt{1+6 x^2}}$.$\frac{x}{\sqrt{1-x^2}} = \frac{x \sqrt{6}}{\sqrt{1+6 x^2}} \Rightarrow 1+6x^2 = 6(1-x^2) = 6-6x^2 \Rightarrow 12x^2 = 5 \Rightarrow x = \frac{1}{2} \sqrt{\frac{5}{3}}$.

List $I$	List $II$
$P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value	$1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$
$Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is	$2$. $\sqrt{2}$
$R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is	$3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is	$4$. $1$

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