Question

Prove the following :
$2{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{31}{17}\right)$

Answer 1

Taking $L.H.S.$
$2{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{31}{17}\right)$
$\because 2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$\therefore L.H.S={\tan }^{-1}\left(\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{1}{1-\frac{1}{4}}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{1}{\frac{3}{4}}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{4}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)$
$(\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy})$

$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{\frac{4\times 7+3\times 1}{7\times 3}}{\frac{7\times 3-4}{7\times 3}}\right)$
$\Rightarrow L.H.S=\left(\frac{\frac{28+3}{21}}{\frac{21-4}{21}}\right)$
$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)$
$\Rightarrow L.H.S={\tan }^{-1}(\frac{31}{21}\times \frac{21}{17})$
$\Rightarrow L.H.S={\tan }^{-1}\left(\frac{31}{17}\right)=R.H.S.$
Hence Proved.