Question

Prove that : ${\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{63}{16}$

Answer 1

Let ${\sin }^{-1}\frac{5}{13}=x\Rightarrow \sin x=\frac{5}{13}\Rightarrow \cos x=\frac{12}{13}$

$\therefore \tan x=\frac{5}{12}\Rightarrow x={\tan }^{-1}\frac{5}{12}$ .........(1)

Let ${\cos }^{-1}\frac{3}{5}=y\Rightarrow \cos y=\frac{3}{5}\Rightarrow \sin y=\frac{4}{5}$

$\therefore \tan y=\frac{4}{3}\Rightarrow y={\tan }^{-1}\frac{4}{3}$ .......(2)

Using (1) and (2), we have
$\text{L.H.S.}={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}$

$={\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{4}{3}$

$={\tan }^{-1}\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times \frac{4}{3}}\right)$

$={\tan }^{-1}\left(\frac{15+48}{36-20}\right)$

$={\tan }^{-1}\frac{63}{16}=\text{R.H.S.}$