Question

If $0\le \alpha ,\beta \le {90}^{\circ }$ and $tan(\alpha +\beta )=3$ and $tan(\alpha -\beta )=2$, then the value of $sin2\alpha$ is -

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

Answer 1

Answer: (B)

$\frac{1}{\sqrt{2}}$

Here, $0\le \alpha ,\beta \le {90}^{\circ }\Rightarrow 1^{st}$quadrant.

Let $A=\alpha +\beta$ and $B=\alpha -\beta \Rightarrow A+B=2\alpha$

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

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

$=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )}$

$=\frac{3+2}{1-3\times 2}=\frac{5}{-5}=-1$

$\Rightarrow \tan 2\alpha =-1<0\Rightarrow 2^{nd}$ quadrant.

Now $0\le \alpha \le {90}^{\circ }\Rightarrow 0\le 2\alpha \le {180}^{\circ }\Rightarrow 1^{st}$ or $2^{nd}$ quadrant.

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

$\therefore \sin 2\alpha =\sin (\pi -\frac{\pi }{4})=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$