Prove that $\displaystyle (\operatorname{cosec} \mathrm{A}-\sin \mathrm{A})(\sec \mathrm{A}-\cos \mathrm{A})=\frac{1}{\cot \mathrm{~A}+\tan \mathrm{A}}$.

\[\begin{aligned} & \text { LHS }=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \\ & =\frac{1-\sin ^{2} A}{\sin \mathrm{~A}} \times \frac{1-\cos ^{2} A}{\cos \mathrm{~A}} \\ & =\sin \mathrm{A} \cos \mathrm{~A} \\ & \text { RHS }=\frac{1}{\dfrac{\cos \mathrm{~A}}{\sin \mathrm{~A}}+\dfrac{\sin \mathrm{A}}{\cos \mathrm{~A}}} \\ & =\frac{\sin A \cos A}{\sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}} \\ & =\sin \mathrm{A} \cos \mathrm{~A}=\mathrm{LHS} \end{aligned} \]