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

Let $S_{n} = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + \dots$ to $n^{\text{th}}$ term. Then $\lim_{n \rightarrow \infty} S_{n}$ is

If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$, then $x$ is equal to

$\begin{aligned} & \text{If } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & b>a, \text{ then } x= \end{aligned}$

If $f(n) = \tan \left[\tan ^{-1} \frac{1}{1+2} + \tan ^{-1} \frac{1}{1+6} + \tan ^{-1} \frac{1}{1+12} + \ldots + \tan ^{-1} \frac{1}{1+n(n+1)}\right]$, then $f(2021) =$

Differentiate the function $\cos^{-1}(\sin x)$ with respect to $x$.