Trigonometric indentity proof

Question

How to prove the following identity?

$$\frac{\mathrm{sin}(2a)-\mathrm{sin}(2b)}{\mathrm{sin}(2a)+\mathrm{sin}(2b)}=\frac{\mathrm{tan}(a-b)}{\mathrm{tan}(a+b)}$$

Answer 1

\begin{align} \dfrac{\tan(a-b)}{\tan(a+b)} &= \dfrac{\sin(a-b) \cos(a+b)}{\cos(a-b) \sin(a+b)} \\ &= \dfrac{2 \sin(a-b) \cos(a+b)}{2 \sin(a+b) \cos(a-b)} \\ &= \dfrac{\sin((a-b)+(a+b)) + \sin((a-b)-(a+b))} {\sin((a+b)+(a-b)) + \sin((a+b)-(a-b))} \\ &= \dfrac{\sin(2a) - \sin(2b)} {\sin(2a) + \sin(2b)} \\ \end{align}

Answer 2

Use the following substitution $$\pmatrix{a & b} = \pmatrix{ \frac{c+d}{2} & \frac{c-d}{2} } $$

So now we have:

$$ \begin{align} \sin(2 a) & = \cos(c) \sin(d) + \sin(c) \cos(d) \\ \sin(2 b) & = \sin(c) \cos(d) - \cos(c) \sin(d) \end{align} $$

and

$$ \begin{align} \sin(2 a) - \sin(2 b) & = 2 \cos(c) \sin(d) = 2 \cos(a+b) \sin(a-b) \\ \sin(2 a) + \sin(2 b) & = 2 \sin(c) \cos(d) = 2 \sin(a+b) \cos(a-b) \end{align} $$

Now divide the above to get your identity.