If $\theta = 2 \tan^{-1} \frac{1}{8} + 2 \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}$ and $\tan \frac{\theta}{2} = \sqrt{m} + \sqrt{n}$,where $m$ and $n$ are positive integers such that $m < n$,then $(m^n + n^m)^{m+n}$ is equal to

If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive integers,then the possible values of $(x, y)$ are:

If $\sum_{k=1}^n \tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} ( \theta )$,then $\theta =$

The number of positive integral solutions of $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ are

Show that $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$

Consider the statements:
$(I)$ If $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then $f(0) = \frac{1}{2}$.
$(II)$ $\sin \left(4 \tan ^{-1} \frac{1}{5} - \tan ^{-1} \frac{1}{239}\right) = 1$.
Then the correct option among the following is: