Trigonometric proof involving several identities 
Show that $$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$


I brought everything to the common denominator on the right hand side.  What should I do next?
 A: First the LHS$${1+\sin A\over\cos A}+{\cos B\over 1-\sin B}={\sin{A\over 2}+\cos {A\over 2}\over-\sin{A\over 2}+\cos {A\over 2} }+{\sin{B\over 2}+\cos {B\over 2}\over \sin{B\over 2}-\cos {B\over 2}}={2\cos{A+B\over 2}\over \cos {A-B\over 2}-\sin{A-B\over 2}}$$
Now the  RHS
$${2(\sin A-\sin B)\over \sin(A-B)+\cos A-\cos B }={2(2\sin{A-B\over 2}\cos {A+B\over 2})\over 2\sin{A-B\over 2}\cos{A-B\over 2}-2\sin{A+B\over 2}\sin{A-B\over 2}}={2\cos{A+B\over 2}\over \cos {A-B\over 2}-\sin{A-B\over 2}}$$
Hence PROVED
A: Partly to see if I could do it, here is a mindless slog, offering no insight at all!
Multiplying across and using the obvious abbreviations $sA=\sin A$ etc, we have $$(1+sA)(s(A-B)+cA-cB)(1-sB)+cBcA(s(A-B)+cA-cB)=2(sA-sB)(1-sB)cA$$ and hence
$$\left(s(A-B)+cA-cB\right)(1+sA-sB-sAsB)+cAcBs(A-B)+cA^2cB-cAcB^2-2cAsA+2cAsAsB+2cAsB-2cAsB^2=0$$
Now substitute $s(A-B)=sAcB-cAsB$ to get: $$\left(sAcB-cAsB+cA-cB\right)(1+sA-sB-sAsB)+cAcB(sAcB-cAsB)+cA^2cB-cAcB^2-2cAsA+2cAsAsB+2cAsB-2cAsB^2=0$$ 
It may not be immediately obvious, but the LHS expands to $0$ also. Collecting first terms with four elements, we have $$-sA^2cBsB+cAsAsB^2+cAsAcB^2-cA^2cBsB=cAsA-cBsB$$ Collecting the terms with three elements we have $$sA^2cB-sAcBsB-cAsAsB+cAsB^2-cAsAsB+sAcBsB+cA^2cB-cAcB^2+2cAsAsB-2cAsB^2$$ $$=cB-cAsB^2-cAcB^2=cB-cA$$ Collecting the terms with two elements we have $$sAcB-cAsB+cAsA-cAsB-sAcB+cBsB-2cAsA+2cAsB=-cAsA+cBsB$$ and finally the terms with only one element $$cA-cB$$ Collecting all this together we get $$cAsA-cBsB+cB-cA-cAsA+cBsB+cA-cB=0$$.
Twenty years ago it was still important to be able to do this kind of thing, but today software is much faster and makes far fewer mistakes doing such slogs, so I suspect their day has gone!
Doing it by hand took me nearly 30 mins (because of errors). Mathematica took less than 2 (including typing it in etc):

