Question

If $\frac{\sin \alpha }{a}=\frac{\cos \alpha }{b}$ , then $a\sin 2\alpha =$

• A. $a$
• B. $\frac{2a^{2}b}{a^2+b^2}$
• C. $a+b$
• D. $\frac{2a^{}{b}^2}{a^2+b^2}$

Answer 1

The correct option is B $\frac{2a^{2}b}{a^2+b^2}$

Given$:\frac{\sin \alpha }{a}=\frac{\cos \alpha }{b}$

$\Rightarrow \frac{\sin \alpha }{\cos \alpha }=\frac{a}{b}$

$\Rightarrow \tan \alpha =\frac{a}{b}$

$\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }$

$a\sin 2\alpha =\frac{a\times 2\times \frac{a}{b}}{1+\frac{a^2}{b^2}}$

$=\frac{2a^{2}b}{a^2+b^2}$