# If 0<= aplha,beta <= 90 and tan(alpha+beta)=3 and tan(alpha-beta)=2 then value of sin(2alpha) is?

## A) $\frac{1}{\sqrt{2}}$ B) $- \frac{1}{\sqrt{2}}$ C) $\frac{1}{2}$ D) None of these

Aug 7, 2018

$\sin \left(2 \alpha\right) = \frac{1}{\sqrt{2}}$

#### Explanation:

Here,

$0 \le \alpha , \beta \le {90}^{\circ} \implies {1}^{s t} Q u a \mathrm{dr} a n t$

Let $A = \alpha + \beta \mathmr{and} B = \alpha - \beta \implies A + B = 2 \alpha$

$\therefore \text{ Using } \tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\tan \left(2 \alpha\right) = \tan \left[\left(\alpha + \beta\right) + \left(\alpha - \beta\right)\right]$

tan(2alpha)=(tan(alpha+beta)+tan(alpha-beta))/(1- tan(alpha+beta)tan(alpha-beta))

$\tan \left(2 \alpha\right) = \frac{3 + 2}{1 - 3 \cdot 2} = \frac{5}{- 5}$

$\tan \left(2 \alpha\right) = - 1 < 0$

Now ,

0 <= alpha <= 90^circ=>0 <= 2alpha <= 180^circ=>1^(st) or 2^(nd)Quadrant

But , $\tan \left(2 \alpha\right) = - 1 < 0 \implies {2}^{n d} Q u a \mathrm{dr} a n t$

So , $2 \alpha = \pi - \frac{\pi}{4}$

$\therefore \sin \left(2 \alpha\right) = \sin \left(\pi - \frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right)$

$\therefore \sin \left(2 \alpha\right) = \frac{1}{\sqrt{2}}$