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

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Correct option is C

We have, sin⁡B=15sin⁡(2A+B)= sin⁡(2A+B)sin⁡B=51= sin⁡(2A+B)+sin⁡Bsin⁡(2A+B)−sin⁡B=5+15−1⇒ 2sin⁡(A+B)cos⁡A 2sin⁡Acos⁡(A+B)=32⇒ tan⁡(A+B)tan⁡A=32

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If sin⁡B=15sin⁡(2A+B), then tan⁡(A+B)tan⁡A is equal to

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