Question

If $\tan (\alpha +\beta )=\sqrt{3},\tan (\alpha -\beta )=1$ then $4\beta =$

Answer 1

$\tan (\alpha +\beta )=\sqrt{3}$

$\Rightarrow \alpha +\beta ={\tan }^{-1}\sqrt{3}=\frac{\pi }{3}...\left(1\right)$

$\tan (\alpha -\beta )=1$

$\Rightarrow \alpha -\beta ={\tan }^{-1}1=\frac{\pi }{4}...\left(2\right)$

Adding (1) and (2) we get

$2\alpha =\frac{7\pi }{12}\Rightarrow \alpha =\frac{7\pi }{24}$

subtract (1) and (2)

$2\beta =\frac{\pi }{12}$

$\Rightarrow 4\beta =\frac{\pi }{6}$