Browse Questions

# Prove the following : $\tan^{-1} \bigg(\large\frac{1}{2} \bigg) + \tan^{-1} \bigg( \large\frac{1}{5} \bigg) + \tan^{-1} \bigg( \frac{1}{8} \bigg)=\large\frac{\pi}{4}$

Toolbox:
• $tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg) , xy < 1$
• $tan^{-1}1=\frac{\pi}{4}$
By taking $x=\frac{1}{2}\:and\:y=\frac{1}{5}$ in the formula of $tan^{-1}x+tan^{-1}y$ we get
L.H.S.=
$tan^{-1} \bigg[\large \frac{\large\frac{1}{2}+\large\frac{1}{5}}{1-\large\frac{1}{10}}\bigg] +tan^{-1}\large\frac{1}{8}$
$= tan^{-1}\large\frac{7}{9}+tan^{-1}\large\frac{1}{8}$
Again by taking $x=\frac{7}{9}\:and\:y=\frac{1}{8}$ in the same formula we get
$= tan^{-1} \bigg[\large \frac{\large\frac{7}{9}+\large\frac{1}{8}}{1-\large\frac{7}{72}} \bigg]$
$= tan^{-1}1=\large\frac{\pi}{4}$
=R.H.S.

edited Mar 22, 2013