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

\[\begin{aligned} & \text { LHS }=\frac{\dfrac{\sin \theta}{\cos \theta}}{1-\dfrac{\cos \theta}{\sin \theta}}+\frac{\dfrac{\cos \theta}{\sin \theta}}{1-\dfrac{\sin \theta}{\cos \theta}} \\ & =\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)} \\ & =\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)} \\ & =\frac{\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta}{\sin \theta \cos \theta} \\ & =\frac{1}{\sin \theta \cos \theta}+1 \\ & =1+\operatorname{cosec} \theta \sec \theta=\mathrm{RHS} \end{aligned} \]