The value of $x$, where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is

$\sec ^2(\tan ^{-1} 2) + \operatorname{cosec}^2(\cot ^{-1} 3)$ is equal to

Let $f : R \to R$,$f(x) = \max\{|\tan^{-1}x|, \cot^{-1}x\}$. Consider the following statements:
$I.$ The function is continuous and derivable $\forall x \in R$.
$II.$ The range of the function is $\left[ \frac{\pi}{4}, \pi \right]$.
$III.$ $f(x)$ is many-one into.
Identify the correct option.

$\cot \left\{\frac{2019 \pi}{2}-\left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)\right\}$ is equal to:

The number of real solutions of $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$ is

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 $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is	$3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is	$4$. $1$

Codes: $P \quad Q \quad R \quad S$