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.
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$\begingroup$You might want to use the following: $\sin(A - B) = \sin(A) \cos(B) - \cos(A) \sin(B)$
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$$\frac{1+\sin A}{\cos A}=\frac{\cos A}{1-\sin A}$$
Then,
$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{\cos A}{1-\sin A}+\frac{\cos B}{1-\sin B}=\frac{(\cos A+\cos B)+\sin(A-B)}{(1-\sin A)(1-\sin B)}$$
Multiply the right-hand side by $1=\frac{(\cos A-\cos B)+\sin(A-B)}{(\cos A-\cos B)+\sin(A-B)}$ and simplify. It's messy, but doable.
$\endgroup$ $\begingroup$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 $$ \left(\frac{1 + \sin A}{\cos A} + \frac{1 + \sin B}{\cos B}\right) \left(\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}\right) = \\ = \frac{1-\sin^2 B}{\cos^2 B} - \frac{1-\sin^2 A}{\cos^2 A} + \frac{(1+\sin A)(1-\sin B)-(1-\sin A)(1+\sin B)}{\cos A\cos B} = \\ = \frac{2\sin A - 2\sin B}{\cos A\cos B} = 2\left(\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}\right). \qquad \blacksquare $$
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