Question

If $\sin (\alpha +\beta )=1$,$\sin (\alpha -\beta )=\frac{1}{2}$,then $\tan (\alpha +2\beta )\tan (2\alpha +\beta )=$

• A. $1$
• B. $-1$
• C. Zero
• D. None of thies

Answer 1

The correct option is A $1$

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

$\alpha +\beta =\frac{\pi }{2}-\left(1\right)$

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

$add\left(1\right)and\left(2\right),$

$\alpha +\beta +\alpha -\beta =\frac{\pi }{2}+\frac{\pi }{6}$

$\Rightarrow 2\alpha =\frac{4\pi }{6}$ $\Rightarrow \alpha =\frac{\pi }{3}$

$Puttingvalueof\alpha in\left(1\right)$

$\frac{\pi }{3}+\beta =\frac{\pi }{2}$

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

$\therefore tan(\alpha +2\beta )tan(2\alpha +\beta )=tan(\frac{\pi }{3}+\frac{\pi }{3})tan(\frac{2\pi }{3}+\frac{\pi }{6})$ $=tan\left(\frac{2\pi }{3}\right)tan\left(\frac{5\pi }{6}\right)$ $=(-\sqrt{3})\left(\frac{-1}{\sqrt{3}}\right)=1$