Prove that $\;$ $\dfrac{\left(\cot A - \text{cosec }A\right)^2 + 1}{\sec A \left(\text{cosec }A - \cot A\right)} = 2 \cot A$
$\begin{aligned}
LHS & = \dfrac{\left(\cot A - \text{cosec }A\right)^2 + 1}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\
& = \dfrac{\cot^2 A + \text{cosec}^2 A - 2 \cot A \text{cosec }A + 1}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\
& \left[\text{Note: }1 + \cot^2 \theta = \text{cosec}^2 \theta\right] \\\\
& = \dfrac{\text{cosec}^2 A + \text{cosec}^2 A - 2 \cot A \text{cosec }A}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\
& = \dfrac{2 \; \text{cosec}^2 A - 2 \cot A \text{cosec }A}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\
& = \dfrac{2 \; \text{cosec }A \left(\text{cosec }A - \cot A\right)}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\
& = \dfrac{2 \; \text{cosec }A}{\sec A} \\\\
& = \dfrac{2 / \sin A}{1 / \cos A} \\\\
& = \dfrac{2 \; \cos A}{\sin A} \\\\
& = 2 \cot A = RHS
\end{aligned}$
Hence proved.