Question

Prove that

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

Answer 1

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

We know that,

$2{\tan }^{-1}A={\tan }^{-1}\left(\frac{2A}{1-A^2}\right)$

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

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

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

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

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

We know that

${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

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

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

$={\tan }^{-1}\left(\frac{\frac{31}{21}}{\frac{21-4}{21}}\right)$

$={\tan }^{-1}\left(\frac{31}{17}\right)=$R.H.S

Hence proved.