Prove the identity $\frac{\sin x}{1-\cos x} + \frac{1-\cos x}{\sin x} = 2\csc x.$ Hence solve the equation $\frac{\sin x}{1-\cos x} + \frac{1-\cos x}{\sin x} = 5$ for $0 \le x \le 2\pi$.

1. Combine the fractions on the left-hand side over a common denominator.

$\text{LHS} = \frac{\sin x}{1-\cos x} + \frac{1-\cos x}{\sin x}$
$= \frac{\sin^2 x + (1-\cos x)^2}{\sin x(1-\cos x)}$

2. Expand the numerator.

$= \frac{\sin^2 x + 1 - 2\cos x + \cos^2 x}{\sin x(1-\cos x)}$

3. Apply the Pythagorean identity to simplify the numerator.

Since $\sin^2 x + \cos^2 x = 1$:
$= \frac{1 + 1 - 2\cos x}{\sin x(1-\cos x)}$
$= \frac{2 - 2\cos x}{\sin x(1-\cos x)}$

4. Factorise the numerator, cancel common factors, and apply the reciprocal identity to reach the RHS.

$= \frac{2(1 - \cos x)}{\sin x(1-\cos x)}$
$= \frac{2}{\sin x}$
$= 2\csc x = \text{RHS}$

1. Substitute the identity from part (a) into the equation

From part (a), we have $\frac{\sin x}{1-\cos x} + \frac{1-\cos x}{\sin x} = 2\csc x$.
So the equation becomes:
$2\csc x = 5$

2. Rewrite in terms of sine and find the basic angle

$\csc x = \frac{5}{2}$
$\sin x = \frac{2}{5} = 0.4$
Basic angle $\alpha = \sin^{-1}(0.4) \approx 0.41152$ rad

3. Find all solutions in the given interval $0 \le x \le 2\pi$

Since $\sin x > 0$, $x$ is in the 1st and 2nd quadrants.
$x = 0.41152 \text{ or } \pi - 0.41152$
$x \approx 0.412 \text{ or } 2.73$