First slide

Compound angles

Question

If $\sin B = \frac{1}{5} \sin (2 A + B),$ then $\frac{\tan (A + B)}{\tan A}$ is equal to

Moderate

A

$5 / 3$
B

$2 / 3$
C

$3 / 2$
D

$3 / 5$

Solution

We have,
$\begin{array}{l} \sin B = \frac{1}{5} \sin (2 A + B) \\ = \frac{\sin (2 A + B)}{\sin B} = \frac{5}{1} \\ = \frac{\sin (2 A + B) + \sin B}{\sin (2 A + B) - \sin B} = \frac{5 + 1}{5 - 1} \\ \Rightarrow \frac{2 \sin (A + B) \cos A}{2 \sin A \cos (A + B)} = \frac{3}{2} \end{array}$
$\Rightarrow \frac{\tan (A + B)}{\tan A} = \frac{3}{2}$

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