Question

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

Answer 1

$LHS={\text{tan}}^{-1}(\frac{1}{5})+{\text{tan}}^{-1}(\frac{1}{7})+{\text{tan}}^{-1}(\frac{1}{3})+{\text{tan}}^{-1}(\frac{1}{8})$

$={\text{tan}}^{-1}(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{35}})+{\text{tan}}^{-1}(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{24}})={\text{tan}}^{-1}(\frac{6}{17})+{\text{tan}}^{-1}(\frac{11}{23})$

$={\text{tan}}^{-1}(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{66}{17\times 23}})={\text{tan}}^{-1}\left(1\right)=\frac{\pi }{4}=RHS$

$LHS=RHS$

Hence Proved