Question

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

where $0\le \alpha$, $\beta \le \frac{\pi }{2}thenfindthevaluesof\tan (\alpha +2\beta )andtan(2\alpha +\beta ).$

Answer 1

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

$\Rightarrow \sin (\alpha +\beta )=\sin \frac{\pi }{2}$

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

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

$\Rightarrow \sin (\alpha -\beta )=\sin \frac{\pi }{6}$ or $\sin (\pi -\frac{\pi }{6})$

$\Rightarrow \alpha -\beta =\frac{\pi }{6}$ or $\pi -\frac{\pi }{6}$

$\Rightarrow \alpha -\beta =\frac{\pi }{6}$ or $\frac{5\pi }{6}$

But $\beta \le \frac{\pi }{2}$ and $0\le \alpha$(given)

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

Equations $\left(1\right)+\left(2\right)$ gives

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

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

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

substitute the value of $\alpha =\frac{\pi }{3}$ in $\left(1\right)$ we get

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

$\therefore \alpha =\frac{\pi }{3}$ and $\beta =\frac{\pi }{6}$

Now,

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

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

$=\tan (\pi -\frac{\pi }{3})$

$=-\tan \frac{\pi }{3}=-\sqrt{3}$

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

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

$=\tan (\pi -\frac{\pi }{6})$

$=-\tan \frac{\pi }{6}=-\frac{1}{\sqrt{3}}$