Question

Find the value of the expression :
$\tan ({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})$

Answer 1

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

Putting ${\sin }^{-1}\left(\frac{3}{5}\right)=x$ and ${\cot }^{-1}\left(\frac{3}{2}\right)=y$
Or $\sin \left(x\right)=\frac{3}{5}$ and $\cot y=\frac{3}{2}$
$\sin \left(x\right)=\frac{3}{5}\Rightarrow \cos x=\sqrt{1-{\sin }^{2}x}=\frac{4}{5}$
$\Rightarrow \sec x=\frac{5}{4}$
$\Rightarrow \tan \left(x\right)=\sqrt{{\sec }^{2}x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}$ and
$\tan y=\frac{1}{\cot y}=\frac{2}{3}$
$\Rightarrow \tan x=\frac{3}{4}$ and $\tan y=\frac{2}{3}$

Now,
$\tan ({\sin }^{-1}\left(\frac{3}{5}\right)+{\cot }^{-1}\left(\frac{3}{2}\right))=\tan (x+y)$
$=\frac{\tan x+\tan y}{1-\tan x\cdot \tan y}$
$=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}=\frac{\frac{3\left(3\right)+2\left(4\right)}{4\times 3}}{\frac{4\times 3-3\times 2}{4\times 3}}$

$=\frac{\frac{9+8}{4\times 3}}{\frac{12-6}{4\times 3}}=\frac{\frac{17}{4\times 3}}{\frac{6}{4\times 3}}=\frac{17}{6}$