Evaluate: $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)$

The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

The value of $\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}$ is

Prove that $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$,where $|x| < \frac{1}{\sqrt{3}}$.

Statement-$1$: ${\cot ^{ - 1}}\left[ {\frac{{\log (e/{x^2})}}{{\log (ex^2)}}} \right] + {\cot ^{ - 1}}\left[ {\frac{{\log (ex^2)}}{{\log (e/{x^2})}}} \right] = \frac{\pi}{2}$
Statement-$2$: ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right] = {\tan ^{ - 1}}1 + {\tan ^{ - 1}}(\log {x^2})$

If $\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]+\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]+\cdots+\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right]=\tan ^{-1}[x]$,then $x=$