(P)        Expression = ${(\frac{1}{{\mathrm{y}}^2}{\left(\frac{\frac{1}{\sqrt{1+{\mathrm{y}}^2}}+\frac{{\mathrm{y}}^2}{\sqrt{1+{\mathrm{y}}^2}}}{\frac{\sqrt{1-{\mathrm{y}}^2}}{\mathrm{y}}+\frac{\mathrm{y}}{\sqrt{1-{\mathrm{y}}^2}}}\right)}^2+{\mathrm{y}}^4)}^{\frac{1}{2}}$  = (1 – y⁴ + y⁴)^1/2 = 1

(Q)        (cos x + cos y) = (– cos z)

and       (sin x + sin y) = (– sin z)

 $\therefore$ On squaring and adding, we get

1 + 1 + 2 cos (x – y) = 1

$\Rightarrow \cos (\mathrm{x}-\mathrm{y})=\frac{-1}{2}\Rightarrow 2{\cos }^2\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\frac{-1}{2}\Rightarrow \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)-1=\frac{-1}{2}$

 $\therefore \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\pm \frac{1}{2}$

(R)        $\cos 2\mathrm{x}(\cos (\frac{\mathrm{\pi }}{4}+\mathrm{x})-\cos (\frac{\mathrm{\pi }}{4}-\mathrm{x}))=\sin 2\mathrm{x}(\frac{\mathrm{sinx}}{\mathrm{cosx}}-\frac{\mathrm{cosx}}{\mathrm{sinx}})$

 $\Rightarrow \mathrm{sinx}=\frac{1}{\sqrt{2}}\Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{4}$

 $\therefore \mathrm{secx}=\sqrt{2}$

(S)      $\frac{\left|\mathrm{x}\right|}{\sqrt{1-{\mathrm{x}}^2}}=\frac{\mathrm{x}\sqrt{6}}{\sqrt{1+6{\mathrm{x}}^2}}\Rightarrow 1+6{\mathrm{x}}^2=6-6{\mathrm{x}}^2\Rightarrow \mathrm{x}=\pm \frac{1}{2}\sqrt{\frac{5}{3}}$

But       x > 0

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