Solve - sin2A-Bsin A-2cosA-B-sin Bsin A

LHS= sin(2A+B) sinA 2cos(A+B) = sin2AcosB+cos2AsinB sinA 2[ cosAcosB sinAsinB ] = 2sinAcosAcosB+(1 2 sin ^ 2 A)sinB 2sinAcosAcosB+2 sin ^ ...

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Trigonometric Ratios of Compound Angles

Solve : sin...

Question

Solve : $\frac{sin (2 A + B)}{sin A} - 2 cos (A + B) = \frac{sin B}{sin A}$

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Solution

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

\frac{s i n (2 A + B)}{s i n A} - 2 c o s (A + B) = \frac{s i n B}{s i n A}

\frac{s i n B}{s i n A} = \frac{s i n (2 A + B)}{s i n A} - 2 c o s (A + B)

\frac{\sin (A + B) + \sin (A - B)}{\cos (A + B) + \cos (A - B)} = \tan A

\frac{\sin (A - B)}{\cos A \cos B} + \frac{\sin (B - C)}{\cos B \cos C} + \frac{\sin (C - A)}{\cos C \cos A} = 0

\frac{\sin (A - B)}{\sin A \sin B} + \frac{\sin (B - C)}{\sin B \sin C} + \frac{\sin (C - A)}{\sin C \sin A} = 0

\frac{\tan (A + B)}{\cot (A - B)} = \frac{\tan^{2} A - \tan^{2} B}{1 - \tan^{2} A \tan^{2} B}

\sqrt{2} cos A = cos B + {cos}^{3} B

\sqrt{2} sin A = sin B - {sin}^{3} B

| sin (A - B) |

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