Question

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

Answer 1

$\text{L.H.S.}=2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}$

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

$={\tan }^{-1}\left(\frac{1}{\frac{3}{4}}\right)+{\tan }^{-1}\frac{1}{7}$

$={\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{7}$

$={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}.\frac{1}{7}}\right)$

$={\tan }^{-1}\frac{\left(\frac{28+3}{21}\right)}{\left(\frac{21-4}{21}\right)}$

$={\tan }^{-1}\frac{31}{17}=\text{R.H.S.}$