$\tan \left(\cos ^{-1} \frac{1}{\sqrt{2}}+\tan ^{-1} \frac{1}{2}\right) = $

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) = $

The number of solutions of $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$ is

$\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12))$ is equal to :
(The inverse trigonometric functions take the principal values)

If $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ then $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$

The derivative of $\tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right)$ with respect to $\sin^{-1} \left( \frac{2x}{1+x^2} \right)$ is...