#### Question

Prove that

\[\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin \left( A + B \right)}{\sin \left( A - B \right)}\]

#### Solution

\[LHS = \frac{\tan A + \tan B}{\tan A - \tan B}\]

\[ = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}\]

\[ = \frac{\frac{\sin A \cos B + \cos A\sin B}{\cos A \cos B}}{\frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B}}\]

\[ = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}\]

\[ = \frac{\sin\left( A + B \right)}{\sin\left( A - B \right)} \]

= RHS

Hence proved .

Is there an error in this question or solution?

Solution Prove that Tan a + Tan B Tan a − Tan B = Sin ( a + B ) Sin ( a − B ) Concept: Trigonometric Functions of Sum and Difference of Two Angles.