Prove that $\left(\frac{\tan A}{1-cotA}\right)+\left(\frac{cotA}{1-\tan A}\right)=1+\tan A+cotA$

To prove: $\left(\frac{\tan A}{1-cotA}\right)+\left(\frac{cotA}{1-\tan A}\right)=1+\tan A+cotA$

Consider L.H.S :

$\begin{array}{rcl}\left(\frac{\tan A}{1-cotA}\right)+\left(\frac{cotA}{1-\tan A}\right)& =& \frac{\tan A}{1-{\displaystyle \frac{1}{\tan A}}}+\frac{{\displaystyle \frac{1}{\tan A}}}{1-\tan A}\\ & =& \frac{\tan A}{{\displaystyle \frac{\tan A-1}{\tan A}}}+\frac{1}{\tan A(1-\tan A)}\\ & =& \frac{{\tan }^{2}A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)}\\ & =& \frac{1-{\tan }^{3}A}{\tan A(1-\tan A)}\mathbf{\left[}\mathbf{\because }\mathbf{}{\mathbf{a}}^{\mathbf{3}}\mathbf{-}{\mathbf{b}}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{ab}\mathbf{)}\mathbf{\right]}\\ & =& \frac{(1-\tan A)(1+{\tan }^{2}A+\tan A)}{\tan A(1-\tan A)}\\ & =& \frac{1+{\tan }^{2}A+\tan A}{\tan A}\\ & =& \frac{1}{\tan A}+\frac{{\tan }^{2}A}{\tan A}+\frac{\tan A}{\tan A}\\ & =& cotA+\tan A+1\\ & =& 1+\tan A+cotA\end{array}$

Hence, it is proved that $\left(\frac{\tan A}{1-cotA}\right)+\left(\frac{cotA}{1-\tan A}\right)=1+\tan A+cotA$