Answer
$\dfrac{1-\cos x}{\sin x}+\dfrac{\sin x}{1-\cos x}=2\csc x$
Work Step by Step
$\dfrac{1-\cos x}{\sin x}+\dfrac{\sin x}{1-\cos x}=2\csc x$
Evaluate the sum on the left side of the equation:
$\dfrac{(1-\cos x)^{2}+\sin^{2}x}{(1-\cos x)\sin x}=2\csc x$
$\dfrac{1-2\cos x+\cos^{2}x+\sin^{2}x}{\sin x-\sin x\cos x}=2\csc x$
Replace $\cos^{2}x+\sin^{2}x$ by $1$ and simplify the numerator:
$\dfrac{1-2\cos x+1}{\sin x-\sin x\cos x}=2\csc x$
$\dfrac{2-2\cos x}{\sin x-\sin x\cos x}=2\csc x$
Take out common factor $2$ from the numerator and common factor $\sin x$ from the denominator and simplify:
$\dfrac{2(1-\cos x)}{\sin x(1-\cos x)}=2\csc x$
$\dfrac{2}{\sin x}=2\csc x$
Since $\dfrac{2}{\sin x}=2\cdot\dfrac{1}{\sin x}=2\csc x$, the identity is proved:
$2\csc x=2\csc x$
