$\tan ^{-1} 2 + \tan ^{-1} 3 = $

Solve $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1} \left(\frac{x-y}{x+y}\right)$ is equal to

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1,$ then find the value of $x$.

If $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$,then $\cot ^{-1} x+\cot ^{-1} y$ is equal to

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 $x, y$,$9 x^2-6 x y \cos \alpha+y^2$ is equal to

If $y = \tan^{-1} \frac{x}{1+2x^2} + \tan^{-1} \frac{x}{1+6x^2} + \tan^{-1} \frac{x}{1+12x^2}$,then $\left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = $