Question

Show that ${\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{2}{11}\right)={\tan }^{-1}\left(\frac{3}{4}\right)$

Answer 1

We know that
${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy},xy<1$
Taking $L.H.S$
${\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{2}{11}\right)$
$={\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{2}{11}}{1-\frac{1}{2}\times \frac{2}{11}}\right)$
$={\tan }^{-1}\left(\frac{\frac{1\left(11\right)+2\left(2\right)}{2\times 11}}{\frac{1\left(11\right)-1}{11}}\right)$
$={\tan }^{-1}\left(\frac{\frac{11+4}{2\times 11}}{\frac{11-1}{11}}\right)$
$={\tan }^{-1}\left(\frac{\frac{15}{2\times 11}}{\frac{10}{11}}\right)={\tan }^{-1}\left(\frac{15}{20}\right)$
$={\tan }^{-1}\left(\frac{3}{4}\right)$
Hence Proved.