If $tan (A + B) = 3 tan A,$ then show that $sin (2 A + B) = 2 sin B$ and $sin 2 (A + B) + sin 2 A = 2 sin 2 B$ .

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Solution

$tan (A + B) = 3 tan A \Rightarrow \frac{sin (A + B)}{cos (A + B)} = \frac{3 sin A}{cos A}$
$\Rightarrow sin (A + B) cos A = 3 sin A cos (A + B)$ ...(1)
$\Rightarrow sin (A + B) cos A - sin A cos (A + B) = 2 sin A cos (A + B) \Rightarrow sin B = sin (A + A + B) + sin (A - A - B) \Rightarrow sin B = sin (2 A + B) - sin B$
$\Rightarrow sin (2 A + B) = 2 sin B$ ...(2)
Now
$sin 2 (A + B) + sin 2 A = 2 sin (\frac{2 A + 2 B + 2 A}{2}) cos (\frac{2 A + 2 B - 2 A}{2}) = 2 sin (2 A + B) cos B = 4 sin B cos B = 2 sin 2 B$

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tan (A + B) = 3 tan A

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