Question

Prove that $2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{31}{17}$.

Answer 1

LHS $=2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}$

$={\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{2}}{1-\frac{1}{2}\times \frac{1}{2}}\right)+{\tan }^{-1}\frac{1}{7}$

$\because ({\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right);xy<1)$

LHS $={\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)={\tan }^{-1}\frac{31}{17}$ = R.H.S