For P

$$\text { Let } t=\sqrt{\frac{1}{y^2}\left(\frac{\cos \left(\tan ^{-1} y\right)+y \cdot \sin \left(\tan ^{-1} y\right)}{\cot \left(\sin ^{-1} y\right)+\tan \left(\sin ^{-1} y\right)}\right)^2+y^4}$$

We know that

$$\begin{gathered} \tan ^{-1} y=\cos ^{-1}\left(\frac{1}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) \\ \text { and } \sin ^{-1} y=\cot ^{-1}\left(\frac{\sqrt{1-y^2}}{y}\right)=\tan ^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) \end{gathered}$$

$$\therefore \quad t = \sqrt {{1 \over {{y^2}}}{{\left[ {{{\cos \left( {{{\cos }^{ - 1}}{1 \over {\sqrt {1 + {y^2}} }}} \right) + y\,.\,\sin \left( {{{\sin }^{ - 1}}{y \over {\sqrt {1 + {y^2}} }}} \right)} \over {\cot \left( {{{\cot }^{ - 1}}{{\sqrt {1 - {y^2}} } \over y}} \right) + \tan \left( {{{\tan }^{ - 1}}{y \over {\sqrt {1 - {y^2}} }}} \right)}}} \right]}^2} + {y^4}} $$

$$\begin{aligned} & \Rightarrow t=\sqrt{\frac{1}{y^2} \times\left[\frac{\frac{1}{\sqrt{1+y^2}}+\frac{y^2}{\sqrt{1+y^2}}}{\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}}\right]^2+y^4} \\ & \Rightarrow t=\sqrt{\frac{1}{y^2} \times\left[\frac{\sqrt{1+y^2}}{\frac{1-y^2+y^2}{y \sqrt{1-y^2}}}\right]^2+y^4} \\ & \Rightarrow t=\sqrt{\frac{1}{y^2} \times y^2\left(1+y^2\right)\left(1-y^2\right)+y^4} \\ & \Rightarrow t=\sqrt{\left(1-y^4\right)+y^4}=1 \\ \end{aligned}$$

Hence, P match with 4.

For Q.

Given $$\cos x+\cos y+\cos z=0$$

$$\Rightarrow \cos x+\cos y=-\cos z$$

On squaring both side

$$\Rightarrow \cos ^2 x+\cos ^2 y+2 \cos x \cdot \cos y=\cos ^2 z\quad \text{... (i)}$$

Also given $$\sin x+\sin y+\sin z=0$$

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

On squaring both side

$$\Rightarrow \sin ^2 x+\sin ^2 y+2 \sin x \cdot \sin y=\sin ^2 z \quad \text{... (ii)}$$

Add equation (i) and (ii)

$$\begin{gathered} \Rightarrow\left(\cos ^2 x+\sin ^2 x\right)+\left(\cos ^2 y+\sin ^2 y\right) \\ +2[\cos x \cdot \cos y+\sin x \cdot \sin y] \\ =\cos ^2 z+\sin ^2 z \end{gathered}$$

$$\begin{array}{lr} \Rightarrow & 1+1+2 \cos (x-y)=1 \\ \Rightarrow & \cos (x-y)=\frac{-1}{2} \\ \Rightarrow & 2 \cos ^2\left(\frac{x-y}{2}\right)-1=\frac{-1}{2} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \cos ^2\left(\frac{x-y}{2}\right)=\frac{1}{4} \\ \Rightarrow & \cos \left(\frac{x-y}{2}\right)= \pm \frac{1}{2} \end{array}$$

Hence, Q match with 3.

For R

Given,

$$\begin{aligned} & \cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \cdot \sin 2 x \cdot \sec x \\ & =\cos x \cdot \sin 2 x \cdot \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x \end{aligned}$$

$$\begin{aligned} \Rightarrow & {\left[\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{\pi}{4}+x\right)\right] } \\ & \cos 2 x+(\sin x-\cos x) \sin 2 x \cdot \sec x=0 \end{aligned}$$

$$\begin{aligned} \Rightarrow & 2 \cdot \sin \frac{\pi}{4} \cdot \sin x \cdot \cos 2 x+(\sin x-\cos x) \\ & \frac{2 \sin x \cdot \cos x}{\cos x}=0 \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad \sqrt{2} \cos 2 x=2(\cos x-\sin x) \\ & \Rightarrow \quad\left(\cos ^2 x-\sin ^2 x\right)=\sqrt{2} (\cos x-\sin x) \\ \end{aligned}$$

$$\begin{array}{rlrl} \Rightarrow & \cos x+\sin x =\sqrt{2} \\ \Rightarrow & (\cos x+\sin x)^2 =(\sqrt{2})^2 \\ \Rightarrow & \cos ^2 x+\sin ^2 x+2 \sin x \cdot \cos x =2 \\ \Rightarrow & 1+\sin 2 x =2 \\ \Rightarrow & \sin 2 x =1 \\ \Rightarrow & 2 x =\frac{\pi}{2} \\ \Rightarrow & x =\frac{\pi}{4} \\ \Rightarrow & \sec x =\sec \frac{\pi}{4}=\sqrt{2} \end{array}$$

Hence, R match with 2

For S

Given,

$$\begin{aligned} \cot \left(\sin ^{-1} \sqrt{1-x^2}\right) & =\sin \left(\tan ^{-1}(x \sqrt{6})\right), x \neq 0 \\ \Rightarrow \cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\right) & =\sin \left(\sin ^{-1} \frac{x \sqrt{6}}{\sqrt{1+6 x^2}}\right) \\ \Rightarrow \quad \frac{x}{\sqrt{1-x^2}} & =\frac{x \sqrt{6}}{\sqrt{1+6 x^2}} \end{aligned}$$

$$ \begin{array}{rlrl} \Rightarrow & \frac{1}{\sqrt{1-x^2}} & =\frac{\sqrt{6}}{\sqrt{1+6 x^2}} \\ \Rightarrow & \text { on squaring both sides } \\ \Rightarrow & \frac{1}{1-x^2} =\frac{6}{1+6 x^2} \\ \Rightarrow & 1+6 x^2=6-6 x^2 \\ \Rightarrow & x^2 =\frac{5}{12} \\ \Rightarrow & x= \pm \sqrt{\frac{5}{12}}= \pm \frac{1}{2} \sqrt{\frac{5}{3}} \end{array}$$

Hence, S match with 1.

Hints :

(i) Recall the method of conversion of one inverse trigonometric function into other inverse trigonometric function as

$$\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}$$

(ii) Recall $$\cos \left(\cos ^{-1} x\right)=x, \sin \left(\sin ^{-1} x\right)=x\tan \left(\tan ^{-1} x\right)=x \text { etc }$$

(iii) Recall the following trigonometric identities

(A) $$\sin ^2 x+\cos ^2 x=1$$

(B) $$\sin 2 x=2 \sin x \cdot \cos x$$

(C) $$\cos 2 x=\cos ^2 x-\sin ^2 x=2 \cos ^2 x-1= 1-2 \sin ^2 x$$

(D) $$\cos (A-B)-\cos (A+B)=2 \sin A \cdot \sin B$$