Match List I with List II and select the correct answer using the code given below the lists : | List-I | List-II | | --- | --- | | (P) $\left(\frac{1}{y^{2}}\left(\frac{\cos(\tan^{-1}y) + y\sin(\tan^{-1}y)}{\cot(\sin^{-1}y) + \tan(\sin^{-1}y)}\right)^{2} + y^{4}\right)^{1/2}$ takes value | (1) $\frac{1}{2}\sqrt{\frac{5}{3}}$ | | (Q) If $\cos x + \cos y + \cos z = 0 = \sin x + \sin y + \sin z$ then possible value of $\cos\frac{x-y}{2}$ is | (2) $\sqrt{2}$ | | (R) If $\cos\left(\frac{\pi}{4}-x\right)\cos 2x + \sin x\sin 2x\sec x = \cos x\sin 2x\sec x + \cos\left(\frac{\pi}{4}+x\right)\cos 2x$ then possible value of $\sec x$ is | (3) $\frac{1}{2}$ | | (S) If $\cot\left(\sin^{-1}\sqrt{1-x^{2}}\right) = \sin\left(\tan^{-1}\left(x\sqrt{6}\right)\right), x \neq 0$, then possible value of $x$ is | (4) $1$ |

$\text{List-II: } (1)\ \dfrac{1}{2}\sqrt{\dfrac{5}{3}} \quad (2)\ \sqrt{2} \quad (3)\ \dfrac{1}{2} \quad (4)\ 1$

$P \to\ ?,\quad Q \to\ ?,\quad R \to\ ?,\quad S \to\ ?$

$\text{then choose the correct code}$

$1.\ \text{solve } (P), (Q), (R), (S)$

$2.\ \text{match each to List-II}$

$3.\ \text{select the correct code}$

$\theta = \tan^{-1}y, \qquad \phi = \sin^{-1}y$

$\cos\theta = \dfrac{1}{\sqrt{1+y^2}}, \quad \sin\theta = \dfrac{y}{\sqrt{1+y^2}}$

$\text{Num} = \dfrac{1}{\sqrt{1+y^2}} + \dfrac{y^2}{\sqrt{1+y^2}} = \sqrt{1+y^2}$

$\text{Den} = \dfrac{\sqrt{1-y^2}}{y} + \dfrac{y}{\sqrt{1-y^2}} = \dfrac{1}{y\sqrt{1-y^2}}$

$\dfrac{\text{Num}}{\text{Den}} = \sqrt{1+y^2}\cdot y\sqrt{1-y^2} = y\sqrt{1-y^4}$

$(P) = \sqrt{\dfrac{1}{y^2}\cdot y^2(1-y^4) + y^4} = \sqrt{1} = 1 \Rightarrow P \to (4)$

$e^{ix} + e^{iy} + e^{iz} = 0$

$\Rightarrow \text{three unit vectors summing to zero}$

$\Rightarrow \text{pairwise angles } 120^\circ$

$x - y = \pm 120^\circ$

$\cos\dfrac{x-y}{2} = \cos 60^\circ = \dfrac{1}{2} \Rightarrow Q \to (3)$

$\cos\!\left(\tfrac{\pi}{4}-x\right) = \dfrac{\cos x + \sin x}{\sqrt2}, \quad \cos\!\left(\tfrac{\pi}{4}+x\right) = \dfrac{\cos x - \sin x}{\sqrt2}$

$\left[\cos(\tfrac{\pi}{4}-x) - \cos(\tfrac{\pi}{4}+x)\right]\cos 2x = (\cos x - \sin x)\sin 2x \sec x$

$\sqrt2 \sin x \cos 2x = 2(\cos x - \sin x)\sin x$

$\sin x\left[\sqrt2\cos 2x - 2(\cos x - \sin x)\right] = 0$

$(\cos x - \sin x)(\cos x + \sin x - \sqrt2) = 0 \Rightarrow \sec x = \sqrt2$

$\Rightarrow R \to (2)$

$A = \sin^{-1}\sqrt{1-x^2}: \ \cot A = \dfrac{|x|}{\sqrt{1-x^2}}$

$B = \tan^{-1}(x\sqrt6): \ \sin B = \dfrac{x\sqrt6}{\sqrt{1+6x^2}}$

$\dfrac{|x|}{\sqrt{1-x^2}} = \dfrac{x\sqrt6}{\sqrt{1+6x^2}} \Rightarrow x > 0$

$\dfrac{1}{\sqrt{1-x^2}} = \dfrac{\sqrt6}{\sqrt{1+6x^2}}$

$1 + 6x^2 = 6(1-x^2) \Rightarrow 12x^2 = 5$

$x = \dfrac{1}{2}\sqrt{\dfrac{5}{3}} \Rightarrow S \to (1)$

$P \to (4), \quad Q \to (3), \quad R \to (2), \quad S \to (1)$

$\Rightarrow \text{option (2)}$