Question

Prove that: ${\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{8}=\frac{\pi }{4}$

Answer 1

${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$

$\therefore {\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{5+7}{35-1}={\tan }^{-1}\frac{6}{17}$

${\tan }^{-1}\frac{6}{17}+{\tan }^{-1}\frac{1}{3}={\tan }^{-1}\frac{18+17}{51-6}={\tan }^{-1}\frac{7}{9}$

${\tan }^{-1}\frac{7}{9}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{56+9}{72-7}={\tan }^{-1}1=\frac{\pi }{4}$

Hence proved