Prove that : \[\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{1}{\sin \theta}-\sin \theta\right)=\frac{1}{\tan \theta+\cot \theta} . \]

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