Prove that $\dfrac{\tan 2\theta}{\sec 2\theta - 1} = \cot \theta$.

1. Express the left-hand side in terms of sine and cosine

$\text{LHS} = \frac{\tan 2\theta}{\sec 2\theta - 1} = \frac{\frac{\sin 2\theta}{\cos 2\theta}}{\frac{1}{\cos 2\theta} - 1}$

2. Simplify the complex fraction

$\text{LHS} = \frac{\frac{\sin 2\theta}{\cos 2\theta}}{\frac{1 - \cos 2\theta}{\cos 2\theta}} = \frac{\sin 2\theta}{1 - \cos 2\theta}$

3. Apply double angle formulae for sine and cosine

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

4. Simplify to reach the right-hand side

$\text{LHS} = \frac{2\sin \theta \cos \theta}{2\sin^2 \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta = \text{RHS}$