Question

Prove that $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{(x+y)}{(1-xy)}$

Answer 1

Let us suppose.

$$\begin{gathered} {\tan }^{-1}x=\alpha ,{\tan }^{-1}y=\beta \\ i.e\alpha =\tan xand\beta =\tan y \end{gathered}$$

Now, we know that $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }$

So,

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

Hence Proved.