Question

If $\cos A=n\cos B$ and $\sin A=m\sin B$, then $(m^2-n^2){\sin }^{2}B=$

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

Answer 1

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

Given
$\cos A=n\cos B\Rightarrow {\cos }^{2}A=n^2{\cos }^{2}B\cdots \left(1\right)$
and
$\sin A=m\sin B\Rightarrow {\sin }^{2}A=m^2{\sin }^{2}B\cdots \left(2\right)$
adding both the equation we have
$1=m^2{\sin }^{2}B+n^2{\cos }^{2}B{m}^2{\sin }^{2}B+n^2-n^2{\sin }^{2}B=1(m^2-n^2){\sin }^{2}B=1-n^2$