(cosecA-sinA) (secA-cosA)=1/sin^2A-sin^4A

$$\begin{gathered} To Prove : (\cos ec A-\sin A)(sec A-\cos A)=\frac{1}{{\sin }^{2}A-{\sin }^{4}A} \\ Solution : LHS= \\ \Rightarrow (\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A) \\ \Rightarrow \left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{1-{\cos }^{2}A}{\cos A}\right) \\ \Rightarrow \left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{{\sin }^{2}A}{\cos A}\right) \\ \Rightarrow \frac{{\sin }^{2}A-{\sin }^{4}A}{\sin A \cos A} \\ \Rightarrow \frac{\cancel{\sin A}(\sin A-{\sin }^{3}A)}{\cancel{\sin A} \cos A} \\ \Rightarrow \frac{\sin A-{\sin }^{3}A}{ \cos A} \\ RHS= \\ =\frac{1}{{\sin }^{2}A-{\sin }^{4}A} \\ =\frac{1}{\sin A(\sin A-{\sin }^{3}A)} \\ =\frac{\cos ec A}{\sin A-{\sin }^{3}A} \\ Since LHS\ne RHS \\ So, this identity doesnot hold. so, cannot be proved \end{gathered}$$
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