Question

Prove that :
$\sin {}^{-1}\frac{3}{5}+\cos {}^{-1}\frac{12}{13}=\sin {}^{-1}\left(\frac{56}{65}\right)$

Answer 1

Let,

${\cos }^{-1}\frac{12}{13}=a\Rightarrow \cos a=\frac{12}{13}$

$\therefore \sin a=\frac{5}{13}$

${\sin }^{-1}\frac{3}{5}=b\Rightarrow \sin b=\frac{3}{5}$

$\therefore \cos b=\frac{4}{5}$

$\sin (a+b)=\sin a\cos b+\cos a\sin b$

$=\frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}$

$=\frac{20}{65}+\frac{36}{65}=\frac{56}{65}$

$\sin (a+b)=\frac{56}{65}$

$a+b={\sin }^{-1}\frac{56}{65}$

putting values,

${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\sin }^{-1}\frac{56}{65}$

hence proved