Show that $\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$ Show that
$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$

$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}\\=\frac{\cos A}{1-\sin A}+\frac{\cos B}{1-\sin B} \\
=\frac{\cos A+\cos B-\sin B\cos A-\cos B\sin A}{(1-\sin A)(1-\sin B)}= \ \\
\frac{\cos A+\cos B-\sin(A+B)}{1-\sin A-\sin B+\sin A\sin B}$$
I am stuck here, I could not solve further. Please help.

The $LHS$ can be written as $\dfrac {1+\sin A}{\cos A} +\dfrac{1+\sin B}{\cos B}$
$=\dfrac{\cos A+\cos B+\sin A\cos B+\cos A\sin B}{\cos A\cos B}$
Multiply Neumaretor and Denominator by $\sin(A-B)+\cos A-\cos B=(\cos A+\sin A\cos B)-(\cos B+\cos A\sin B)$
$=\dfrac{(\cos A+\sin A\cos B)^2-(\cos B+\cos A\sin B)^2}{\cos A\cos B[...]}$
$=\dfrac{\cos^2 A+\sin^2 A\cos^2B+2\sin A\cos A\cos B-\cos^2B-\cos^2A\sin^2B-2\cos A\cos B\sin B}{\cos A\cos B[...]}$
Note that $\sin^2A\cos^2B-\cos^2A\sin^2B=(1-\cos^2A)\cos^2B-\cos^2A(1-\cos^2B)=\cos^2B-\cos^2A$
So LHS becomes
$\dfrac{2\sin A \cos A\cos B-2\cos A\cos B\sin B}{\cos A\cos B[...]}$
$=\dfrac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}=RHS$
