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$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=$
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The correct answer is:
$\tan A+\cot A+1$
$\begin{aligned} & \text {} \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} \\ & =\frac{\sin ^2 A}{\cos A(\sin A-\cos A)}+\frac{\cos ^2 A}{\sin A(\cos A-\sin A)} \\ & =\frac{1}{\sin A-\cos A}\left[\frac{\sin ^2 A}{\cos A}-\frac{\cos ^2 A}{\sin A}\right] \\ & =\frac{\sin ^3 A-\cos ^3 A}{(\sin A-\cos A) \cos ^2 A \sin A} \\ & =\frac{(\sin A-\cos A)\left(\cos ^2 A+\sin ^2 A+\sin A \cos A\right)}{(\sin A-\cos A) \cos A \cdot \sin A} \\ & =\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}+1=\cot A+\tan A+1\end{aligned}$
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