Question

Prove that: ${\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)=\left(\frac{\pi }{2}\right)$.

Answer 1

Let $\tan \alpha =\frac{1}{2}$ and $\tan \beta =\frac{1}{3}$

$\therefore {\tan }^{-1}1+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\frac{\pi }{4}+\alpha +\beta$

Hence it suffices to prove that ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\alpha +\beta =\frac{\pi }{4}$.

Now, consider

$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$⟹\tan (\alpha +\beta )=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}$

$⟹\tan (\alpha +\beta )=\frac{\frac{5}{6}}{\frac{5}{6}}$

$⟹\tan (\alpha +\beta )=1$

$⟹\alpha +\beta =\frac{\pi }{4}$ (taking inverse on both sides)

$⟹{\tan }^{-1}1+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\frac{\pi }{2}$ (adding ${\tan }^{-1}1$ on both sides)

Hence proved.