Question

Prove that ${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{(x+y)}{(1-xy)}$ when xy<1.

Answer 1

$\to$ Let $\alpha =ta{n}^{-1}x,\beta =ta{n}^{-1}y$

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

$\Rightarrow \tan (\alpha +\beta )=\frac{x+y}{1-xy}$

$\Rightarrow \alpha +\beta ={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=ta{n}^{-1}x+ta{n}^{-1}y$