Prove that : $\displaystyle (1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)=2$

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