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
 A: You did everything thus far correctly, I just pick up with where you left off in the second line:
$$\begin{align}(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A) 
& = \dfrac{1 - 2 \cos A + \cos^2 A + \sin^2 A}{\sin A(1 - \cos A)} \\ \\
& = {1 \color{blue}{\bf + \cos^2 A} -2\cos A + 1 \color{blue}{\bf - \cos^2A} \over \sin A(1-\cos A)} \\ \\
& = \dfrac{2 - 2\cos A}{\sin A(1 - \cos A)}\\ \\
& = \dfrac{2\color{red}{\bf (1-\cos A)}}{\sin A\color{red}{\bf (1 - \cos A)}}\\ \\
& = \frac{2}{\sin A} \\ \\
& = 2 \csc A
\end{align}$$
A: Hint:
$1-\cos A=1-(1-2 \sin^2\dfrac{A}{2})=2\sin^2 \dfrac{A}{2}$
$\dfrac{2 \sin^2 \dfrac{A}{2}}{2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}}=\tan \dfrac{A}{2}$
The other expression will be $\cot \dfrac{A}{2}$
$(\tan^2 \dfrac{A}{2}+1) /\tan\dfrac{A}{2}= \dfrac{\sec^2 A \cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}= \dfrac{1}{2 \sin A}$
A: $$ LHS =\frac {1 - \cos A} {\sin A} + \frac {\sin A} {1 - \cos A} $$
$$ = \frac {2 \sin^2 \frac A2} {2\sin \frac A2 \cos \frac A2} + \frac {2\sin \frac A2 \cos \frac A2}{2 \sin^2 \frac A2}$$
$$ = \frac {\sin \frac A2} {\cos \frac A2} + \frac {\cos \frac A2} {\sin \frac A2} $$
Now just cross multiply and you get the answer.
A: When we "cross-multiply," on top we get $(1-\cos A)^2+\sin^2 A$. Expand the square.
We get $1-2\cos A+\cos^2 A+\sin^2 A$. Replace $\cos^2 A+\sin^2 A$ by $1$. We get $2-2\cos A$ on top, and now it's over.  
A: 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:
$$ 2 \csc A$$
A: LCM is not required.
Observe that the first term already has $\sin A$ in the denominator.
$$\text{Now,}\frac{\sin A}{1-\cos A}=\frac{\sin A(1+\cos A)}{1-\cos^2A}=\frac{\sin A(1+\cos A)}{\sin^2A}=\frac{1+\cos A}{\sin A}$$
$$\text{So,}{1- \cos A \over \sin A } + { \sin A \over 1- \cos A} =\frac{1-\cos A}{\sin A}+\frac{1+\cos A}{\sin A}=\frac2{\sin A}=2\csc A$$
