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

Then \(\tan x = \frac{s}{c}\).

Substitute into the left-hand side (LHS):

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

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

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

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

Factor the denominator:

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

Cancel the common factor \((1-s)\):

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

This matches the right-hand side (RHS):

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