Question

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

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

Answer 1

The correct option is B $1-n^2$

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
$1=n^2{\cos }^{2}B+m^2{\sin }^{2}B$
$\Rightarrow 1=n^2(1-{\sin }^{2}B)+m^{2}si{n}^{2}B$
Thus $(m^2-n^2){\sin }^{2}B=1-n^2$