Question

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

List - I

P $\quad\left(\frac{1}{y^2}\left(\frac{\cos \left(\tan ^{-1} y\right)+y \sin \left(\tan ^{-1} y\right)}{\cot \left(\sin ^{-1} y\right)+\tan \left(\sin ^{-1} y\right)}\right)^2+y^4\right)^{1 / 2}$ takes value

Q. If $\cos x+\cos y+\operatorname{cox} z=0=\sin x+\sin y+\sin z$ then

R. If $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+$

S. If $\cot \left(\sin ^{-1} \sqrt{1-x^2}\right)=\sin \left(\tan ^{-1}(x \sqrt{6})\right), x \neq 0$,

List - II

1. $\frac{1}{2} \sqrt{\frac{5}{3}}$

2. $\sqrt{2}$ possible value of $\cos \frac{x-y}{2}$ is

3. $\frac{1}{2}$ $\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$ then possible value of $\sec x$ is

4. 1

Then possible value of $x$ is

Codes:

$\begin{array}{lllll} & \mathrm{P} & \mathrm{Q} & \mathrm{R} & \mathrm{S} \\ \text { (A) } & 4 & 3 & 1 & 2 \\ \text { (B) } & 4 & 3 & 2 & 1 \\ \text { (C) } & 3 & 4 & 2 & 1 \\ \text { (D) } & 3 & 4 & 1 & 2\end{array}$

Answer 1

(P) $ \left(\frac{1}{y^2}\left(\frac{\cos \left(\tan ^{-1} y\right)+y \sin \left(\tan ^{-1} y\right)}{\cot \left(\sin ^{-1} y\right)+\tan \left(\sin ^{-1} y\right)}\right)^2+y^4\right. \\$

$=  {\left[\frac{1}{y^2}\left[\frac{\left(\frac{1}{\sqrt{1+y^2}}+\frac{y \cdot y}{\sqrt{1+y^2}}\right)}{\left(\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}\right)}\right]^2+y^4\right]^{1 / 2} } \\$

$=  \left(\frac{1}{y^2} \cdot y^2\left(1-y^4\right)+y^4\right)^{1 / 2}=1$ Ans. 4

(Q)$\cos x+\cos y=-\cos z \\$

$\sin x+\sin y=-\sin z \\$

$2+2 \cos (x-y)=1 \\$

$\Rightarrow \quad \cos (x-y)=-1 / 2 \\$

$\Rightarrow \quad 2 \cos ^2\left(\frac{x-y}{2}\right)-1=-1 / 2, \quad \Rightarrow \cos \left(\frac{x-y}{2}\right)=1 / 2$ square and add Ans. 3

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

$\cos 2 x(\sqrt{2} \sin x)+2 \sin ^2 x=2 \sin x \cos x \Rightarrow \sqrt{2} \sin x[\cos 2 x+\sqrt{2} \sin x-\sqrt{2} \cos x]=0$

Either $\sin x=0$ OR $\cos ^2 x-\sin ^2 x=\sqrt{2}(\cos x-\sin x)$

$\sec x=1 \text { OR } \cos x=\sin x \Rightarrow \sec x=\sqrt{2}$

Ans. 2 

(S) $\quad \cot \left(\sin ^{-1} \sqrt{1-x^2}\right)=\sin \left(\tan ^{-1}(x \sqrt{6})\right)$

$\frac{x}{\sqrt{1-x^2}}=\frac{x \sqrt{6}}{\sqrt{1+6 x^2}} \Rightarrow 1+6 x^2=6-6 x^2 \Rightarrow 12 x^2=5$

$x=\sqrt{\frac{5}{12}}==\frac{1}{2} \sqrt{\frac{5}{3}}$

Ans. 1 

Hence Option B is Correct