<s>[INST] Proving a little tough trigonometric identity Show that $$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$ How do I get the $A-B$ term in the denominator? Is RHS to LHS easier? Thanks.
 [/INST] First, notice that
$$
\frac{1 + \sin A}{\cos A} = \frac{\cos A}{1 - \sin A}
$$
Next, regroup the denomitator as
$$
\sin(A - B) + \cos A - \cos B = \cos A(1 - \sin B) - \cos B(1 - \sin A)
$$
The right hand side becomes
$$
2\frac{\sin A - \sin B}{\cos A(1 - \sin B) - \cos B(1 - \sin A)} = 
2\frac{\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}}{\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}}.
$$
Multiplying both sides with ${\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}}$ we get
$$
\left(\frac{1 + \sin A}{\cos A} + 
\frac{1 + \sin B}{\cos B}\right) 
\left(\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}\right)
=
2\left(\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}\right)
$$
Expanding the left side gives
$$
\left(\frac{1 + \sin A}{\cos A} + 
\frac{1 + \sin B}{\cos B}\right) 
\left(\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}\right) = \\ =
\frac{1-\sin^2 B}{\cos^2 B} - \frac{1-\sin^2 A}{\cos^2 A} 
+ \frac{(1+\sin A)(1-\sin B)-(1-\sin A)(1+\sin B)}{\cos A\cos B} = \\ =
\frac{2\sin A - 2\sin B}{\cos A\cos B} = 2\left(\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}\right). \qquad \blacksquare
$$
</s>