Question

If $\frac{\cos A}{\cos B}=n, \frac{\sin A}{\sin B}=m$, then the value of $\left(m^{2}-n^{2}\right) \sin ^{2} B$ is

• A. $1 + n^2$
• B. $1 - n^2$
• C. $ n^2$
• D. $- n^2$

Answer 1

Answer: (B)

Given, $\frac{\cos A}{\cos B}=n, \frac{\sin A}{\sin B}=m$
Therefore, $\cos A = n \cos B ; \sin A = m \sin B$
Squaring and adding, we get
$1=n^{2} \cos ^{2} B +m^{2} \sin ^{2} B$
$\Rightarrow 1=n^{2}\left(1-\sin ^{2} B\right)+m^{2} \sin ^{2} B$
Thus $\left(m^{2}-n^{2}\right) \sin ^{2} B=1-n^{2}$