STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

Then

$\sin x=\frac{3}{5}$

$\Rightarrow \cos x=\sqrt{1-\sin ^{2} x}=\frac{4}{5}$

$\Rightarrow \sec x=\frac{5}{4}$

$\therefore \tan x=\sqrt{\sec ^{2} x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}$

$\therefore x=\tan ^{-1} \frac{3}{4}$

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

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

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

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

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

$=\tan \left(\tan ^{-1} \frac{9+8}{12-6}\right)$

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