Question

Prove ${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{8}{19}=\frac{\pi }{4}$.

Answer 1

Formula to be used:

${\tan }^{-1}a+{\tan }^{-1}b+{\tan }^{-1}c={\tan }^{-1}\left[\frac{a+b+c-abc}{1-ab-bc-ca}\right]$

${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{8}{19}={\tan }^{-1}\left[\frac{\frac{3}{4}+\frac{3}{5}+\frac{8}{19}-\frac{3}{4}\frac{3}{5}\frac{8}{19}}{1-\frac{3}{4}\frac{3}{5}-\frac{3}{5}\frac{8}{19}-\frac{8}{19}\frac{3}{4}}\right]$

${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{8}{19}={\tan }^{-1}\left[\frac{\frac{425}{380}}{\frac{425}{380}}\right]$

${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{8}{19}={\tan }^{-1}1$

${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{8}{19}=\frac{\pi }{4}$