If $z=\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$ then 6z is equal to _________

17

$z=\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$

so  $\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$

using triangle

$\cot ^{-1} \frac{3}{2} = \tan ^{-1} \frac{2}{3}$

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

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

$\Rightarrow z=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}$

$\Rightarrow z=\frac{9+8}{12-6}=\frac{17}{6}$  so  $6 z=17$

Option: C