Prove that $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$

Q. Prove: $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$}


A. $$\begin{array}{lll}\\
\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)} &= \frac{\frac{1}{tan(x)}}{1 - tan(x)} + \frac{tan(x)}{1 - \frac{1}{tan(x)}}&\mbox{Expressing } cot(x)=1/tan(x)\\
&= \frac{1}{tan(x)(1-tan(x))} + \frac{\tan^2(x)}{tan(x) - 1}& \mbox{Simplifying fractions.}\\
&= \frac{1}{tan(x)(1-tan(x))} - \frac{\tan^2(x)}{1- tan(x)}& \mbox{Last term as a subtraction }\\
&= \frac{1 - tan^3(x)}{tan(x)(1-tan(x)}& \mbox{Rewriting as a single fraction}\\
&= \frac{(1 - tan(x)){(1 + tan(x)} + \tan^2(x)}{tan(x)(1-tan(x))}& \mbox{Factoring a difference of cubes}\\
&= \frac{(1 + tan(x) +\tan^2(x)}{tan(x)}&\mbox{Cancellation by division}\\
&= cot(x) + 1 + tan(x)& \mbox{Division}\\
&= 1 + tan(x) + cot(x)& \mbox{Rearranging terms.}\\
\end{array}
$$