Question

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

Answer 1

Proof:

Let,${\tan }^{-1}x=\alpha$ and ${\tan }^{-1}y=\beta$

Thus, we can say; $x=\tan \alpha ,y=\tan \beta$

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

i.e

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

Hence proved.