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