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

Answer: (B)

$\cos \, A = n \cos B$ and $\sin \, A = m \, \sin \, B$
Squaring and adding, we get
$1 = n^2 \, \cos^2 B + m^2 \sin^2 B$
$\Rightarrow \, 1 = n^2 (1 - \sin^2 B) + m^2 \sin^2 B$
$\therefore \, (m^2 - n^2) \sin^2 B = 1 - n^2$