Question

Prove that ${\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13}-{\tan }^{-1}\frac{2}{9}=0$

Answer 1

Simplifying the LHS of ${\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13}-{\tan }^{-1}\frac{2}{9}=0$

${\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13}-{\tan }^{-1}\frac{2}{9}={\tan }^{-1}\left(\frac{\frac{1}{7}+\frac{1}{13}}{1-\frac{1}{7}\times \frac{1}{13}}\right)-{\tan }^{-1}\frac{2}{9}$

$={\tan }^{-1}\left(\frac{20}{90}\right)-{\tan }^{-1}\left(\frac{2}{9}\right)$

$={\tan }^{-1}\left(\frac{2}{9}\right)-{\tan }^{-1}\left(\frac{2}{9}\right)$

$=0$

$=RHS$

Hence proved.