Taken from my answer to “Proving a weird trig identity” at Math.StackExchange.com .
$$\Large{\frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta + \tan\theta}{1}}$$
In the diagram, \(\overline{AB}\) is tangent to the unit circle at P. The “trig lengths” (except for \(|\overline{AQ}|\)) should be clear.
We note that \(\angle BPR \cong \angle RPP^\prime\), since these inscribed angles subtend congruent arcs \(\stackrel{\frown}{PR}\) and \(\stackrel{\frown}{RP^\prime}\). Very little angle chasing gives that \(\triangle APQ\) is isosceles, with \(\overline{AP}\cong \overline{AQ}\) (justifying that last trig length). Then,
$$\triangle SPR \sim \triangle OQR \Longrightarrow \frac{|\overline{SP}|}{|\overline{SR}|}=\frac{|\overline{OQ}|}{|\overline{OR}|}\Longrightarrow \frac{\cos\theta}{1−\sin\theta}=\frac{\sec\theta+\tan\theta}{1}$$