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

Let $A = {\sin ^{ - 1}}\frac{3}{5}$ and $B = {\cot ^{ - 1}}\frac{3}{2}$
$\Rightarrow$ $\sin A = \frac{3}{5}\;and\;\,\cot B = \frac{3}{2} \Rightarrow$ $\tan A = \frac{3}{4}\;\,and\;\;\tan B = \frac{2}{3}$

So, $\tan \left( {{{\tan }^{ - 1}}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3}} \right)$
$= \tan \left( {{{\tan }^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4} \times \frac{2}{3}}}} \right)} \right) = \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{{9 + 18}}{{12}}}}{{1 - \frac{1}{2}}}} \right)} \right]$

$= \tan \left( {{{\tan }^{ - 1}}\left( {\frac{{17}}{{16}}} \right)} \right) = \frac{{17}}{6}$