Prove that

$$\begin{gathered} \text{LHS=}\frac{\tan A}{(1-\cot A)}+\frac{\cot A}{(1-\tan A)} \\ \text{=}\frac{\tan A}{(1-\cot A)}+\frac{{\cot }^{2}A}{(\cot A-1)}[\because \tan A=\frac{1}{\cot A}] \\ \text{=}\frac{\tan A}{(1-\cot A)}-\frac{{\cot }^{2}A}{(1-\cot A)} \\ \text{=}\frac{\tan A-{\cot }^{2}A}{(1-\cot A)} \\ \text{=}\frac{\left(\frac{1}{\cot A}\right)-{\cot }^{2}A}{(1-\cot A)} \\ \text{=}\frac{1-{\cot }^{3}A}{\cot A(1-\cot A)} \\ \text{=}\frac{(1-\cot A)(1+\cot A+{\cot }^{2}A)}{\cot A(1-\cot A)}[\because a^3-b^3=(a-b)(a^2+ab+b^2)] \\ \text{=}\frac{1}{\cot A}+\frac{{\cot }^{2}A}{\cot A}+\frac{\cot A}{\cot A} \\ \text{=1+}\tan A+\cot A \\ \text{=RHS} \\ \text{Hence proved} \end{gathered}$$