Question

Prove that $\frac{\tan A+\tan B}{\tan A-\tan B}=\frac{\sin \left(A+B\right)}{\sin \left(A-B\right)}$ .

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\tan A+\tan B}{\tan A-\tan B} \\ =\frac{{\displaystyle \frac{\sin A}{\cos A}}+{\displaystyle \frac{\sin B}{\cos B}}}{{\displaystyle \frac{\sin A}{\cos A}}-{\displaystyle \frac{\sin B}{\cos B}}} \\ =\frac{{\displaystyle \frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}}}{{\displaystyle \frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}}} \\ =\frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B} \\ =\frac{\sin \left(A+B\right)}{\sin \left(A-B\right)} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$