Start with the left-hand side:

\(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right)^2\)

\(= \left( \frac{1 - \cos x}{\sin x} \right)^2\)

Let \(s = \sin x\) and \(c = \cos x\). Then:

\(\left( \frac{1 - c}{s} \right)^2 = \frac{(1 - c)^2}{s^2}\)

Using the identity \(s^2 = 1 - c^2\), we have:

\(\frac{(1 - c)^2}{1 - c^2}\)

Factor the denominator:

\(= \frac{(1 - c)(1 - c)}{(1 - c)(1 + c)}\)

Cancel \((1 - c)\) from numerator and denominator:

\(= \frac{1 - c}{1 + c}\)

Thus, the identity is proved:

\(\frac{1 - \cos x}{1 + \cos x}\)