$\cos ^{-1} \frac{3}{5} + \sin ^{-1} \frac{5}{13} + \tan ^{-1} \frac{16}{63} = $

If $x = \sin (2 \tan^{-1} 2)$,$y = \cos (2 \tan^{-1} 3)$,and $z = \sec (2 \tan^{-1} 4)$,then:

$50 \tan \left(3 \tan ^{-1}\left(\frac{1}{2}\right)+2 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1}(2 \sqrt{2})\right)$ is equal to

If $\sinh (2 \tanh ^{-1} x) = \frac{11}{60}$,then $x =$

If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the incorrect option is

If $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$,where $-1 \le x \le 1$,$-2 \le y \le 2$,and $x \le \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to