Find the values of
Question:

Find the values of $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$

Solution:

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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