Prove identity problem

Hello, Jason!

Quote:

Prove: . $\frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x}\; = \;2\csc x$

I've tried several things, to no avail.

I don't suppose you tried adding those two fractions . . .

$\frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x} \;=\;{\color{blue}\frac{1-\cos x}{1-\cos x}}\cdot\frac{1-\cos x}{\sin x} + {\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin x}{1-\cos x}$

. . $= \;\frac{(1-\cos x)^2 + \sin^2x}{\sin x(1-\cos x)}\;=\;\frac{1 - 2\cos x + \overbrace{\cos^2x + \sin^2x}^{\text{This is 1}}}{\sin x(1 - \cos x)}$

. . $= \;\frac{2 - 2\cos x}{\sin x(1 - \cos x)} \;=\;\frac{2(1-\cos x)}{\sin x(1-\cos x)} \;=\;\frac{2}{\sin x} \;=\;2\csc x$