Let \(c = \cos \theta\) and \(s = \sin \theta\). The left-hand side (LHS) is:

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

Expanding the numerator:

\((1 + c)^2 = 1 + 2c + c^2\)

\((1 - c)^2 = 1 - 2c + c^2\)

Thus,

\((1 + c)^2 - (1 - c)^2 = 1 + 2c + c^2 - (1 - 2c + c^2) = 4c\)

The denominator is:

\((1 - c)(1 + c) = 1 - c^2\)

So,

\(\frac{4c}{1 - c^2}\)

Since \(1 - c^2 = s^2\), we have:

\(\frac{4c}{s^2}\)

Using \(\tan \theta = \frac{s}{c}\), we get:

\(\frac{4}{s \cdot \frac{s}{c}} = \frac{4}{s^2/c} = \frac{4c}{s^2}\)

Thus, the identity is proven as \(\frac{4}{s \cdot \frac{s}{c}} = \frac{4}{s \tan \theta}\).