Verify the identity $\frac{\tan(a+b)}{\tan(a-b)}$ = $\frac{\sin(a)\cos(a)+\sin(b)\cos(b)}{\sin(a)\cos(a)-\sin(b)\cos(b)}$

Question

I've been asked to verify the following identity but I don't know how to do it.

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

When I try I get

$$\frac{\tan(a+b)}{\tan(a-b)} = \frac{\dfrac{\sin(a+b)}{\cos(a+b)}}{\dfrac{\sin(a-b)}{\cos(a-b)}} = \frac{\dfrac{\sin(a)\cos(b)+\cos(a)\sin(b)}{\cos(a)\cos(b)-\sin(a)\sin(b)}}{\dfrac{\sin(a)\cos(b)-\cos(a)\sin(b)}{\cos(a)\cos(b)+\sin(a)\sin(b)}}$$ But I don't know really where to go from here.

Answer 1

Hint :

Use $$2\sin x\cos x=\sin2x$$

Then Prosthaphaeresis Formulas $$\sin C-\sin D=?\text{ and }\sin C+\sin D=?$$

Answer 2

This can be somewhat lengthy and can take more time but give it a try :

In the last step you know what you have to apply.

Hope it helps.

Answer 3

Use $$\sin{a}\cos{a}+\sin{b}\cos{b}=\frac{1}{2}(\sin2a+\sin2b)=\sin(a+b)\cos(a-b)$$ and $$\sin{a}\cos{a}-\sin{b}\cos{b}=\frac{1}{2}(\sin2a-\sin2b)=\sin(a-b)\cos(a+b)$$