Question

Prove that : ${\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{2}$.

Answer 1

To prove :

$si{n}^{-1}\frac{4}{5}+si{n}^{-1}\frac{5}{13}+si{n}^{-1}\frac{16}{65}=\frac{\pi }{2}$

LHS :

$si{n}^{-1}\frac{4}{5}+si{n}^{-1}\frac{5}{13}+si{n}^{-1}\frac{16}{65}$

$=ta{n}^{-1}\frac{4}{3}+ta{n}^{-1}\frac{5}{12}+ta{n}^{-1}\frac{16}{63}$

$=ta{n}^{-1}\frac{63}{16}+ta{n}^{-1}\frac{16}{63}$

$=ta{n}^{-1}\frac{63}{16}+co{t}^{-1}\frac{63}{16}$

$=\frac{\pi }{2}$ = RHS