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

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Solution

$\begin{matrix} tan (A + B) = 3 tan A \frac{sin (A + B)}{cos (A + B)} = \frac{3 sin A}{cos A} sin (A + B) cos A = 3 sin A cos (A + B) sin (A + B) cos A - sin A (A + A + B) + sin (A - A - B) sin (A + B - A) = sin (A + A + B) + sin (A - A - B) sin B = sin (2 A + B) - sin B 2 sin B - sin (2 A + B) n o t L H S sin (2 A + 2 B) + sin 2 A 2 sin (2 A + B) G B 2 \times 2 sin B G B = 2 sin 2 B \end{matrix}$

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Similar questions

tan (A + B) = 3 tan A,

sin (2 A + B) = 2 sin B

sin 2 (A + B) + sin 2 A = 2 sin 2 B

If tan(A+B) =3tanA

then show that sin(2A +2B)+sin2A=2sin²B

\frac{1 + sin 2 A - cos 2 A}{1 + sin 2 A + cos 2 A} = tan A

\frac{sin 2 (A + B) + sin 2 B}{sin 2 A} = 2,

tan (A + B)

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