Question

If $\sin (\alpha +\beta )=1,\sin (\alpha -\beta )=\frac{1}{2}$, then $\tan (\alpha +2\beta )\tan (2\alpha +\beta )$ is equal to $\alpha ,\beta ϵ(0,\pi /2)$

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

Answer 1

Answer: (A)

$1$

$\sin (\alpha +\beta )=1\Rightarrow \cos (\alpha +\beta )=0\Rightarrow \cos 3(\alpha +\beta )=0.........\left(1\right)$

$\sin (\alpha +\beta )=\frac{1}{2}\Rightarrow \cos (\alpha -\beta )=\frac{\sqrt{3}}{2}.......\left(2\right)$

Now,

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

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

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

Using $2\sin X\sin Y=\cos (X-Y)-\cos (X+Y)2\cos X\cos Y=\cos (X+Y)+\cos (X-Y)$

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

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

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

Substituting values from $\left(1\right)$ and $\left(2\right)$, we get

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

$=\frac{\frac{\sqrt{3}}{2}+0}{0+\frac{\sqrt{3}}{2}}=1$