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Match List i with List II and select the correct answer using the code given below the lists

A) .P:4,Q:2,R:3,S:1

B) P:4,Q:3,R:2,S:1

C) P:3,Q:4,R:2,S:1

D) P:3,Q:4,R:1,S:2

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

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Answer:

Explanation: