Question

Prove that :

${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}={\sin }^{-1}\frac{1}{\sqrt{5}}$

Answer 1

LHS :

${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}$

$={\tan }^{-1}\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}.\frac{2}{9}}$

$={\tan }^{-1}\frac{9+8}{36-2}={\tan }^{-1}\frac{1}{2}={\sin }^{-1}\frac{1}{\sqrt{5}}$