Question

prove that ${tan}^{-1}\left(\frac{1}{3}\right)+{tan}^{-1}\left(\frac{1}{5}\right)+{tan}^{-1}\left(\frac{1}{7}\right)+{tan}^{-1}\left(\frac{1}{8}\right)=\frac{\pi }{4}$

Answer 1

LHS :

$ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{5}+ta{n}^{-1}\frac{1}{7}+ta{n}^{-1}\frac{1}{8}$

$=ta{n}^{-1}\frac{11}{23}+ta{n}^{-1}\frac{12}{34}$ $[ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{x+y}{1-xy}]$

$=ta{n}^{-1}\frac{11}{23}+ta{n}^{-1}\frac{6}{17}$

$=ta{n}^1\frac{325}{325}$

$=ta{n}^{-1}1$

$=\frac{\pi }{4}$ = RHS