$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}=$

The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals

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

Considering only the principal values of the inverse trigonometric functions,the value of $\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^2}+\tan ^{-1} \frac{\sqrt{2}}{\pi}$ is:

If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$ and $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$,then $2 x^2+y^2-x y=$