Question

If ${\tan }^{-1}\alpha +{\tan }^{-1}\beta =\frac{\pi }{4}$, then twice write the value of $\alpha +\beta +\alpha \beta .$

Answer 1

$ta{n}^{-1}\left(\alpha \right)+ta{n}^{-1}\left(\beta \right)=\frac{\pi }{4}$

$\Rightarrow ta{n}^{-1}\left(\frac{\alpha +\beta }{1-\alpha \beta }\right)=\frac{\pi }{4}$

$\Rightarrow \frac{\alpha +\beta }{1-\alpha \beta }=tan\left(\frac{\pi }{4}\right)=1$

$\Rightarrow \alpha +\beta =1-\alpha \beta$

$\Rightarrow \alpha +\beta +\alpha \beta =1$

$\therefore \alpha +\beta +\alpha \beta =1$ and $2(\alpha +\beta +\alpha \beta )=2$