Question

Show that $ta{n}^{-1}\frac{63}{16}=si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}$

Answer 1

To prove :

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

R.H.S: ${\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{4}{3}$

$[{\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}]$

$={\tan }^{-1}\left[\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times \frac{4}{3}}\right]={\tan }^{-1}\left[\frac{\frac{5+16}{12}}{1-\frac{5}{9}}\right]$

$={\tan }^{-1}\frac{\frac{21}{12}}{\frac{4}{9}}={\tan }^{-1}\frac{63}{16}=$ L.H.S