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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
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Verified Answer
The correct answer is:
$1-\mathrm{n}^{2}$
$\cos A=n \cos B$ and $\sin A=m \sin B$
Squaring and adding, we get $1=n^{2} \cos ^{2} B+m^{2} \sin ^{2} B$
$$
\begin{array}{ll}
\Rightarrow & 1=n^{2}\left(1-\sin ^{2} B\right)+m^{2} \sin ^{2} B \\
\therefore & \left(m^{2}-n^{2}\right) \sin ^{2} B=1-n^{2}
\end{array}
$$
Squaring and adding, we get $1=n^{2} \cos ^{2} B+m^{2} \sin ^{2} B$
$$
\begin{array}{ll}
\Rightarrow & 1=n^{2}\left(1-\sin ^{2} B\right)+m^{2} \sin ^{2} B \\
\therefore & \left(m^{2}-n^{2}\right) \sin ^{2} B=1-n^{2}
\end{array}
$$
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