Question

If A and B are the values such that $(A + B)$ and $(A - B)$ are not odd multiples of $\frac{\pi}{2}$ and $2\tan(A+B) = 3 \tan(A-B)$, then $\sin A \cos A =$

• A. $\sin B \cos B$
• B. $5 \sin B \cos B$
• C. $\sin 2B$
• D. $\cos 2B$

Hint:

Apply tangent addition formulas and manipulate algebraically to relate products of sines and cosines.

Answer 1

The Correct Option is B

Given $2 \tan(A+B) = 3 \tan(A-B)$. Using tangent addition formula:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},
\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}. \] Substitute and cross-multiply. Simplify the resulting equation to express $\sin A \cos A$ in terms of $\sin B \cos B$. After simplification, the relation is $\sin A \cos A = 5 \sin B \cos B$.