Question

If $\sin \left(\alpha +\beta \right)=1$ and $\sin \left(\alpha -\beta \right)=\frac{1}{2}$ where $0\le \alpha ,\beta \le \frac{\pi }{2}$ then find the value of $\tan \left(\alpha +2\beta \right)$ $\tan \left(2\alpha +\beta \right)$

Answer 1

$\sin (\alpha +\beta )=1$

$\alpha +\beta =\pi /2$

$\sin (\alpha -\beta )=1/2$

$\alpha -\beta =\pi /6$

$2\alpha =\frac{2\pi }{3}$

$\alpha =\frac{\pi }{3}$

$\beta =\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}$

$\tan (\alpha +2\beta )\tan (2\alpha +\beta )$

$=\tan (\frac{\pi }{3}+\frac{\pi }{3})\tan (\frac{2\pi }{3}+\frac{\pi }{6})$

$=\tan \left(\frac{2\pi }{3}\right)\tan \left(\frac{5\pi }{6}\right)$

$=(-\sqrt{3})(-1/\sqrt{3})$

$=1$