Another trigonometric proof...? ...sigh..another problem how shall I prove the following?
$$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$
so what now? the following's what I've done:
$$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$

$$
\begin{align}
\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}&=\frac{\cos A\cot A - \sin A \tan A}{\cos A-\sin A}
\\\\&=\frac{\frac{\cos^2 A}{\sin A} - \frac{\sin^2 A}{\cos A}}{\cos A-\sin A}
\\\\&=\frac{\cos^3 A - \sin^3 A}{(\cos A \sin A)(\cos A-\sin A)}
\\\\&=\frac{\cos^2 A + \sin^2 A + \sin A\cos A}{\cos A \sin A}
\\\\&=1+\cot A+\tan A
\end{align}
$$
