Prove that: $(\frac{1}{\cos\theta} – \cos \theta)$ x $(\frac{1}{\sin\theta} – \sin \theta) = \frac{1}{(\tan\theta + \cot \theta)}$

Let’s start from LHS LHS = $(\frac{1}{\cos\theta} – \cos \theta)$ x $(\frac{1}{\sin\theta} – \sin \theta)$ = $(\frac{1 - \cos^2\theta}{\cos \theta})$ x $(\frac{1 - \sin^2\theta}{\sin\theta})$ = $(\frac{\sin^2\theta}{\cos\theta})$ x $(\frac{\cos^2\theta}{\sin\theta})$ = sin θ x cos θ Now, take RHS RHS = $\frac{1}{(\tan\theta + \cot \theta)}$ = $\frac{1}{\frac{\sin\theta|}{\cos\theta} + \frac{\cos \theta}{\sin\theta}}$ = $\frac{1}{\frac{\sin ^2 \theta| + \cos ^2\theta}{\sin \theta \cos\theta}}$ = sin θ x cos θ Since LHS = RHS Hence proved!