Question

If $\sin \left(\alpha +\beta \right)=1,\sin \left(\alpha -\beta \right)=\frac{1}{2},$ $\text{then}\tan \left(\alpha +2\beta \right)\tan \left(2\alpha +\beta \right)$ is equal to

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The correct option is A 1

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

then $\tan (\alpha +2\beta )\tan (2\alpha +\beta )=\frac{\sin (\alpha +2\beta )\sin (2\alpha +\beta )}{\cos (\alpha +2\beta )\cos (2\alpha +\beta )}$

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

$=\frac{\cos (\alpha -\beta )-\cos \left(3\right(\alpha +\beta \left)\right)}{\cos \left(3\right(\alpha +\beta \left)\right)+\cos (\alpha -\beta )}$

by $\left(1\right)$ and $\left(2\right)$

$\cos (\alpha +\beta )=0$ and $\cos (\alpha -\beta )=\sqrt{3}/2$.

$\therefore \tan (\alpha +2\beta )\tan (2\alpha +\beta )=\frac{\sqrt{3}/2-\cos \left(3\right(\alpha +\beta \left)\right)}{\cos \left(3\right(\alpha +\beta \left)\right)+\sqrt{3}/2}$

$=\frac{\sqrt{3}/2-4{\cos }^3(\alpha +\beta )+3\cos (\alpha +\beta )}{4{\cos }^3(\alpha +\beta )-3\cos (\alpha +\beta )+\sqrt{3}/2}$

$=1$