Question

If $\sin \alpha =A\sin (\alpha +\beta ),A\ne 0,$ then the value of $\tan \alpha$ is:

• A. $\frac{A\sin \beta }{1-A\cos \beta }$
• B. $\frac{A\sin \alpha }{1+A\cos \beta }$
• C. $\frac{A\cos \beta }{1-A\sin \beta }$
• D. $\frac{A\sin \alpha }{1-A\cos \beta }$

Answer 1

Answer: (A)

$\frac{A\sin \beta }{1-A\cos \beta }$

$\sin \alpha =A\sin (\alpha +\beta )$

$\Rightarrow \sin \alpha =A\sin \alpha \cos \beta +A\cos \alpha \sin \beta$

$\Rightarrow \sin \alpha (1-A\cos \beta )=A\cos \alpha \sin \beta$

$\Rightarrow \frac{\sin \alpha }{\cos \alpha }=\frac{A\sin \beta }{(1-A\cos \beta )}$

$\Rightarrow \tan \alpha =\frac{A\sin \beta }{(1-A\cos \beta )}$