<s>[INST] How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
 [/INST] I'll go step by step.
$${1 - \cos A \over \sin A} + {\sin A \over 1 - \cos A}$$
$${(1 - \cos A)^2 \over \sin A (1 - \cos A)} + {\sin^2 A \over \sin A(1 - \cos A)}$$
$$(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A)$$
Expanding $(1 - \cos A)^2$ yields:
$$1 - 2\cos A + \cos^2 A + \sin^2 A \over \sin A(1 - \cos A)$$
Knowing that $\cos^2 x + \sin^2 x = 1$, we can change the expression to the following:
$$1 - 2 \cos A + 1 \over \sin A(1 - \cos A)$$
$$2 - 2 \cos A \over \sin A(1 - \cos A)$$
We factorize using $2$:
$$2(1 - \cos A) \over \sin A(1 - \cos A)$$
$$2 \over \sin A$$
$$2 \left ({1 \over \sin A} \right)$$
Finally:
$$ 2 \csc A$$
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