Find the values of $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$
Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-\sin ^{2} x}=\frac{4}{5} \Rightarrow \sec x=\frac{5}{4}$.
$\therefore \tan x=\sqrt{\sec ^{2} x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}$
$\therefore x=\tan ^{-1} \frac{3}{4}$
$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$....(1)
Now, $\cot ^{-1} \frac{3}{2}=\tan ^{-1} \frac{2}{3}$...(2) $\left[\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\right]$
Hence, $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$
$=\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$ [Using (i) and (ii) ]
$=\tan \left(\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)$ $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$
$=\tan \left(\tan ^{-1} \frac{9+8}{12-6}\right)$
$=\tan \left(\tan ^{-1} \frac{17}{6}\right)=\frac{17}{6}$
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