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

\[\begin{aligned} \mathrm{LHS} & =\frac{\dfrac{\sin \theta}{\cos \theta}}{\dfrac{(\sin \theta-\cos \theta)}{\sin \theta}}+\frac{\dfrac{\cos \theta}{\sin \theta}}{\dfrac{(\cos \theta-\sin \theta)}{\cos \theta}} \\ & =\frac{1}{(\sin \theta-\cos \theta)}\left[\frac{\sin ^{2} \theta}{\cos \theta}-\frac{\cos ^{2} \theta}{\sin \theta}\right] \\ & =\frac{1}{(\sin \theta-\cos \theta)} \times \frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta} \\ & =\frac{1}{\sin \theta \cos \theta}+1 \\ & =1+\sec \theta \operatorname{cosec} \theta=\mathrm{RHS} \end{aligned} \]