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$\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}=$
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Verified Answer
The correct answer is:
$\sec A \operatorname{cosec} A+1$
Since, $\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}$
$\begin{aligned}
& =\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}+\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} \\
& =\frac{\cos ^2 A}{\sin A(\cos A-\sin A)}+\frac{\sin ^2 A}{\cos A(\sin A-\cos A)} \\
& =\frac{(\cos A-\sin A)(1+\sin A \cos A)}{(\cos A-\sin A) \sin A \cdot \cos A} \\
& =\sec A \cdot \operatorname{cosec} A+1
\end{aligned}$
$\begin{aligned}
& =\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}+\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} \\
& =\frac{\cos ^2 A}{\sin A(\cos A-\sin A)}+\frac{\sin ^2 A}{\cos A(\sin A-\cos A)} \\
& =\frac{(\cos A-\sin A)(1+\sin A \cos A)}{(\cos A-\sin A) \sin A \cdot \cos A} \\
& =\sec A \cdot \operatorname{cosec} A+1
\end{aligned}$
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