Question

If tan A = x tan B, prove that $\frac{\sin \left(A-B\right)}{\sin \left(A+B\right)}=\frac{x-1}{x+1}$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\sin (A-B)}{\sin (A+B)} \\ =\frac{\sin A\cos B-cosA\sin B}{\sin A\cos B+\cos A\sin B} \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\cos A\cos B: \\ \frac{\tan A-\tan B}{\tan A+\tan B} \\ =\frac{x\tan B-\tan B}{x\tan B+\tan B}(\mathrm{Since}\tan A=x\tan B) \\ =\frac{\tan B\left(x-1\right)}{\tan B\left(x+1\right)} \\ =\frac{x-1}{x+1} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$