Question

If $2\tan A = 3\tan B,$ then $\frac{\sin 2B}{5 - \cos 2B}$ is equal to

• A. tanA– tanB
• B. $\tan(A - B)$
• C. $\tan(A + B)$
• D. $\tan(A + 2B)$

Answer 1

Answer: (B)

$\tan(A - B)$

Answer 2

$2\tan A = 3\tan B$⇒ $\tan A = \frac{3}{2}\tan B = \frac{3}{2}t$

(Let $\tan B = t$) ⇒ $\sin 2B = \frac{2t}{1 + t^{2}},\cos 2B = \frac{1 - t^{2}}{1 + t^{2}}$

∴ $\frac { \sin 2 B } { 5 - \cos 2 B } =$ $\frac{\left( \frac{2t}{1 + t^{2}} \right)}{5 - \left( \frac{1 - t^{2}}{1 + t^{2}} \right)} = \frac{2t}{4 + 6t^{2}} = \frac{t}{2 + 3t^{2}} = \tan(A - B)$.