Match List I with List II and select the correct answer using the...

\left(\text{P}\right) - 4 \text{;} \left(\text{Q}\right) - 3 \text{;} \left(\text{R}\right) - 2 \text{or} 4 \text{;} \left(\text{S}\right) - 1<br />\left(\text{P}\right) {\left(\dfrac{1}{{y}^{2}} {\left(\dfrac{\text\cos \left({\text\tan}^{-1} y\right) + y \text\sin \left({\text\tan}^{-1} y\right)}{\text{cot} \left({\text\sin}^{-1} y\right) + \text\tan \left({\text\sin}^{-1} y\right)}\right)}^{2} + {y}_{4}\right)}^{1 / 2}
$= {\left[\dfrac{1}{{y}^{2}} {\left[\dfrac{\left(\dfrac{1}{\sqrt{1 + {y}^{2}}} + \dfrac{y \text{.} y}{\sqrt{1 + {y}^{2}}}\right)}{\left(\dfrac{\sqrt{1 - {y}^{2}}}{y} + \dfrac{y}{\sqrt{1 - {y}^{2}}}\right)}\right]}^{2} + {y}^{4}\right]}^{1 / 2}$
$= \dfrac{{\text{y}}^{2}}{{\text{y}}^{2}} \left(1 - {\text{y}}^{2}\right) + {\text{y}}^{4}$
$= 1$

4
\left(\text{Q}\right) \text\cos x + \text\cos y = - \text\cos z
\text\sin x + \text\sin y = - \text\sin z \text{Square and add}
2 + 2 \text\cos \left(x - y\right) = 1
⇒ \text\cos \left(x - y\right) = - 1 / 2
⇒ 2 {\text\cos}^{2} \left(\dfrac{x - y}{2}\right) - 1 = - 1 / 2 \text{,} ⇒ \text\cos \left(\dfrac{x - y}{2}\right) = 1 / 2
3
\left(\text{R}\right) \text\cos 2 x \left(\text\cos \left(\dfrac{\pi}{4} - x\right) - \text\cos \left(\dfrac{\pi}{4} + x\right)\right) + 2 {\text\sin}^{2} x = 2 \text\sin x \text\cos x$<br />$\text\cos 2 x \left(\sqrt{2} sin x\right) + 2 {\text\sin}^{2} x = 2 \text\sin x \text\cos x$<br />$\sqrt{2} \text\sin x \left[\text\cos 2 x + \sqrt{2} \text\sin x - \sqrt{2} \text\cos x\right] = 0$<br />Either sin x = 0 OR${\text\cos}^{2} x - {\text\sin}^{2} x = \sqrt{2} \left(\text\cos x - \text\sin x\right)$<br />sec x = 1 OR cos x = sin x<br />$⇒ \text{sec} x = \sqrt{2}<br />2 OR 4<br />\left(\text{S}\right) \text{cot} \left({\text\sin}^{-1} \sqrt{1 - {x}^{2}}\right) = \text\sin \left({\text\tan}^{-1} \left(x \sqrt{6}\right)\right)

$\dfrac{x}{\sqrt{1 - {x}^{2}}} = \dfrac{x \sqrt{6}}{\sqrt{1 + 6 {x}^{2}}}$
$⇒ 1 + 6 {x}^{2} = 6 - 6 {x}^{2}$
$⇒ 1 2 {x}^{2} = 5$
$⇒ x = \sqrt{\dfrac{5}{1 2}} = = \dfrac{1}{2} \sqrt{\dfrac{5}{3}}$
1