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