Question
Prove that:
$
\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\sec A \operatorname{cosec} A
$
$
\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\sec A \operatorname{cosec} A
$
✓
Answer
$\begin{aligned} LHS & =\frac{\frac{\sin A}{\cos A}}{\frac{(\sin A-\cos A)}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{\frac{(\cos A-\sin A)}{\cos A}} \\ & =\frac{1}{(\sin A-\cos A)}\left[\frac{\sin ^2 A}{\cos A}-\frac{\cos ^2 A}{\sin A}\right] \\ & =\frac{1}{(\sin A-\cos A)} \times \frac{(\sin A-\cos A)\left(\sin ^2 A+\cos ^2 A+\sin A \cos A\right)}{\sin A \cos A} \\ & =\frac{1}{\sin A \cos A}+1 \\ & =1+\sec A \operatorname{cosec} A= RHS \end{aligned}$
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