Find the value of $\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)$.

If $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha$,then $\alpha=$

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 $x=$

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-$2$: ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right] = {\tan ^{ - 1}}1 + {\tan ^{ - 1}}(\log {x^2})$

Evaluate: ${\cot ^{ - 1}}3 + {\csc ^{ - 1}}\sqrt 5 = $

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$,where $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,and $x \leq \frac{y}{3}$,then for all $x, y$,$9x^2 - 6xy \cos \alpha + y^2$ is equal to