Question

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

• A. \(1+n^2\)
• B. \(1-n^2\)
• C. \(n^2\)
• D. \(-n^2\)

Hint:

Use \(\sin^2x+\cos^2x=1\) wisely.

Answer 1

The Correct Option is B

Step 1: \[ m^2-n^2=\frac{\sin^2A}{\sin^2B}-\frac{\cos^2A}{\cos^2B} \]
Step 2: \[ =\frac{\sin^2A\cos^2B-\cos^2A\sin^2B}{\sin^2B\cos^2B} =\frac{\sin^2(A-B)}{\sin^2B\cos^2B} \]
Step 3: Using identities, simplifies to \(1-n^2\).