Proving $\frac{\cos A - \cos B}{\sin A + \sin B} = \frac{\sin B - \sin A}{\cos A + \cos B}$ 
Prove $$\dfrac{\cos A - \cos B}{\sin A + \sin B} = \dfrac{\sin B - \sin A}{\cos A + \cos B}$$

I tried as shown below and am not sure how to do it. Your help is appreciated. Thanks.
Proving from left hand side:
$$\dfrac{\cos A}{\sin A + \sin B} - \dfrac{\cos B}{\sin A + \sin B}$$
$$=\dfrac{\dfrac{\cos A}{\sin A}} {\dfrac{\sin A+\sin B}{\sin A}} - \dfrac{\dfrac{\cos B}{\sin B}}{\dfrac{\sin A+\sin B}{\sin B}}$$
$$= \dfrac{\dfrac{1}{\tan A}}{1+\dfrac{\sin B}{\sin A}}- \dfrac{\dfrac{1}{\tan B}}{1+\dfrac{\sin A}{\sin B}}$$

We have that (for $\sin A + \sin B \neq 0$ and $\cos A + \cos B \neq 0$)
$$\dfrac{\cos A - \cos B}{\sin A + \sin B} = \dfrac{\sin B - \sin A}{\cos A + \cos B} $$
$$\iff (\cos A - \cos B)(\cos A + \cos B)=(\sin B - \sin A)(\sin B + \sin A)$$
$$\iff \cos^2 A - \cos^2 B=\sin^2 B - \sin^2 A$$
$$\iff \cos^2 A + \sin^2 A= \cos^2 B + \sin^2 B$$

Edit
As an alternative, by sum to product identities we have

*

*LHS

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

*

*RHS

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