Prove that :
$\frac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) = \cosec A(\cot A - 1)$
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Solving L.H.S,
$$\begin{aligned}\Rightarrow \frac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) \\[1em] \Rightarrow \frac{\sin A(1 - \sin^2 A)}{\cos A(\cos^2 A - 1)} \times \Big(\frac{1}{\cos A} - \frac{1}{\sin A}\Big) \\[1em] \Rightarrow \frac{\sin A(1 - \sin^2 A)}{-\cos A(1- \cos^2 A)} \times \Big(\frac{1}{\cos A} - \frac{1}{\sin A}\Big) \\[1em] \Rightarrow \frac{\sin A (\cos^2 A)}{-\cos A(\sin^2 A)} \times \Big(\frac{\sin A - \cos A}{\cos A \sin A}\Big) \\[1em] \Rightarrow -\frac{\cos A}{\sin A} \times \Big(\frac{\sin A - \cos A}{\cos A \sin A}\Big) \\[1em] \Rightarrow -\Big(\frac{\sin A - \cos A}{\sin^2 A}\Big) \\[1em] \Rightarrow \frac{\cos A - \sin A}{\sin^2 A} \\[1em] \Rightarrow \frac{1}{\sin A} \Big(\frac{\cos A - \sin A}{\sin A}\Big) \\[1em] \Rightarrow \frac{1}{\sin A} \Big(\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A}\Big) \\[1em] \Rightarrow \cosec A (\cot A - 1).\end{aligned}$$
Since,
L.H.S = R.H.S
Hence, proved that
$\frac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) = \cosec A(\cot A - 1)$.