# Find the values of each of the expressions in Exercises 16 to 18.    18. $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

D Divya Prakash Singh

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

we can take $\sin^{-1}\frac{3}{5} = x$,

then  $\sin x = \frac{3}{5}$

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

$\Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

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

We have similarly;

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

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

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

$=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ]$            from   $As, \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$

$=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}$

Exams
Articles
Questions