Question

If $\sin A=n\sin (A+2B)$, then $\frac{\tan \left(B\right)}{\tan (A+B)}$ is equal to,

• A. $\frac{2n}{n+1}$
• B. $\frac{2n}{n-1}$
• C. $\frac{n}{n+1}$
• D. $\frac{1-n}{n+1}$

Answer 1

The correct option is D $\frac{1-n}{n+1}$

We have,

$\sin A=n\sin (A+2B)$

$\sin (A+2B)=\frac{\sin A}{n}$

Now,

$\sin (A+2B)+\sin A=2\sin \frac{(A+2B+A)}{2}\cos \frac{(A+2B-A)}{2}$

$\sin (A+2B)+\sin A=2\sin (A+B)\cos B$

$\frac{\sin A}{n}+\sin A=2\sin (A+B)\cos B$

$\sin A(\frac{1}{n}+1)=2\sin (A+B)\cos B$ $............\left(1\right)$

$\sin (A+2B)-\sin A=2\sin \frac{(A+2B-A)}{2}\cos \frac{(A+2B+A)}{2}$

$\sin (A+2B)-\sin A=2\cos (A+B)\sin B$

$\frac{\sin A}{n}-\sin A=2\cos (A+B)\sin B$

$\sin A(\frac{1}{n}-1)=2\cos (A+B)\sin B$ $............\left(2\right)$

On dividing (1) by (2), we get

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

$\frac{(1-n)}{(1+n)}=\frac{\tan \left(B\right)}{\tan (A+B)}$

Hence, this is the answer.