Question

If $\sin (\alpha +\beta )=1,\sin (\alpha -\beta )=\frac{1}{2}$, $\alpha ,\beta \in [0,\frac{\pi }{2}]$, then

• A. $\tan (\alpha +2\beta )=-\sqrt{3}$
• B. $\tan (2\alpha +\beta )=-\frac{1}{\sqrt{3}}$
• C. $\tan (\alpha -\beta )=1$
• D. None of these

Answer 1

The correct option is A $\tan (\alpha +2\beta )=-\sqrt{3}$

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

$\alpha +\beta =\frac{\pi }{2}$ .......(1)

$\sin (\alpha -\beta )=\frac{1}{2}$

$\alpha -\beta =\frac{\pi }{6}$ .......(2)

Adding equation (1) and (2)

$2\alpha =\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}$

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

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

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

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

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

$=-\sqrt{3}$