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Question: If \(\sin B\) =\(\frac{1}{5}\sin(2A + B)\), then \(\frac{\tan(A + B)}{\tan A} =\)...

If sin⁡B\sin B =15sin⁡(2A+B)\frac{1}{5}\sin(2A + B), then tan⁡(A+B)tan⁡A=\frac{\tan(A + B)}{\tan A} =

A

53\frac{5}{3}

B

23\frac{2}{3}

C

32\frac{3}{2}

D

35\frac{3}{5}

Answer

32\frac{3}{2}

Explanation

Solution

sin⁡(2A+B)sin⁡B=51\frac{\sin(2A + B)}{\sin B} = \frac{5}{1} by componendo and Dividendo.

sin⁡(2A+B)+sin⁡Bsin⁡(2A+B)−sin⁡B=5+15−1\frac{\sin(2A + B) + \sin B}{\sin(2A + B) - \sin B} = \frac{5 + 1}{5 - 1}

2sin⁡(A+B).cos⁡A2cos⁡(A+B).sin⁡A=64\frac{2\sin(A + B).\cos A}{2\cos(A + B).\sin A} = \frac{6}{4} ⇒ tan⁡(A+B)tan⁡A=32\frac{\tan(A + B)}{\tan A} = \frac{3}{2}.