Question

Prove: $\frac{\sin 2A+\sin 2B}{\sin 2A-\sin 2B}=\frac{\tan \left(A+B\right)}{\tan \left(A-B\right)}$.

Answer 1

Given, $\frac{\sin 2A+\sin 2B}{\sin 2A-\sin 2B}=\frac{\tan \left(A+B\right)}{\tan \left(A-B\right)}$

We know that

$$\begin{gathered} \sin \left(A+B\right)=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) \\ \sin \left(A-B\right)=2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right) \end{gathered}$$

Now, consider

$$\begin{gathered} \sin 2A+\sin 2B=2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right) \\ \sin 2A+\sin 2B=2\sin \left(A+B\right)\cos \left(A-B\right) \end{gathered}$$

Similarly

$$\begin{gathered} \sin 2A-\sin 2B=2\sin \left(\frac{2A-2B}{2}\right)\cos \left(\frac{2A+2B}{2}\right) \\ \sin 2A-\sin 2B=2\sin \left(A-B\right)\cos \left(A+B\right) \end{gathered}$$

$$\begin{gathered} \therefore \frac{\sin 2A+\sin 2B}{\sin 2A-\sin 2B}=\frac{2\sin \left(A+B\right)\cos \left(A-B\right)}{2\sin \left(A-B\right)\cos \left(A+B\right)} \\ =\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)}\times \frac{\cos \left(A-B\right)}{\sin \left(A-B\right)} \\ =\tan \left(A+B\right)\times cot\left(A-B\right) \\ =\frac{\tan \left(A+B\right)}{\tan \left(A-B\right)} \end{gathered}$$

Hence proved.