Prove that $\;$ $\left(\text{cosec} \theta - \sin \theta\right) \left(\sec \theta - \cos \theta\right) \left(\tan \theta + \cot \theta\right) = 1$
$\begin{aligned}
LHS & = \left(\text{cosec} \theta - \sin \theta\right) \left(\sec \theta - \cos \theta\right) \left(\tan \theta + \cot \theta\right) \\\\
& = \left(\dfrac{1}{\sin \theta} - \sin \theta\right) \left(\dfrac{1}{\cos \theta} - \cos \theta\right) \left(\tan \theta + \dfrac{1}{\tan \theta}\right) \\\\
& = \left(\dfrac{1 - \sin^2 \theta}{\sin \theta}\right) \left(\dfrac{1 - \cos^2 \theta}{\cos \theta}\right) \left(\dfrac{1 + \tan^2 \theta}{\tan \theta}\right) \\\\
& = \dfrac{\cos^2 \theta}{\sin \theta} \times \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sec^2 \theta}{\tan \theta} \\\\
& = \dfrac{\cos \theta \times \sin \theta}{\cos^2 \theta \times \tan \theta} \\\\
& = \dfrac{\sin \theta}{\cos \theta} \times \dfrac{1}{\tan \theta} \\\\
& = \dfrac{\tan \theta}{\tan \theta} = 1 = RHS
\end{aligned}$
Hence proved.