$\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$ is equal to

If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$, then $\cos(2 \csc^{-1}\sqrt{k - 1}) = \dots$

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 $f(2021) =$

Considering only the principal values of the inverse trigonometric function,the value of $\tan \left(\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right)$ is

The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$,where $-1 \leq x \leq 1, -2 \leq y \leq 2, x \leq \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to