Question

If $\frac{sin\alpha }{a}=\frac{cos\alpha }{b}$ then show that $asin2\alpha +bcos2\alpha =$

• A. b
• B. a
• C. a+b
• D. a-b

Answer 1

The correct option is A b

We have,

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

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

Since,

$=a\sin 2\alpha +b\cos 2\alpha$

$=2a\sin \alpha \cos \alpha +b(1-2{\sin }^2\alpha )$

$=\frac{2a^2{\cos }^2\alpha }{b}+b(1-\frac{2a^2{\cos }^2\alpha }{b^2})$

$=\frac{2a^2{\cos }^2\alpha }{b}+b-\frac{2a^2{\cos }^2\alpha }{b}$

$=b$

Hence, this is the answer.