Question

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

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

Answer 1

The correct option is B $\frac{\sin \alpha (1-A\cos \beta )}{A\cos \alpha \cos \beta }$

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

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

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

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

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

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