Question
Prove that : $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\sec A \cdot \operatorname{cosec} A+1$.
✓
Answer
$\text { LHS }=\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}$
$=\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A-\cos A}+\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A-\sin A}$
$=\frac{\sin ^2 A}{\cos A(\sin A-\cos A)}+\frac{\cos ^2 A}{\sin A(\cos A-\sin A)}$
$=\frac{\sin ^2 A}{\cos A(\sin A-\cos A)}-\frac{\cos ^2 A}{\sin A(\sin A-\cos A)}$
$=\frac{\sin ^3 A-\cos 3 A}{\sin A \cdot \cos A(\sin A-\cos A)}$
$=\frac{1+\sin A \cdot \cos A}{\sin A \cdot \cos A}$
$=\frac{1}{\sin A \cdot \cos A}+\frac{\sin A \cdot \cos A}{\sin A \cdot \cos A}$
$= \sec A.\ce{cosec} \ A + 1$
$= \ce{RHS}$
Hence proved.
$=\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A-\cos A}+\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A-\sin A}$
$=\frac{\sin ^2 A}{\cos A(\sin A-\cos A)}+\frac{\cos ^2 A}{\sin A(\cos A-\sin A)}$
$=\frac{\sin ^2 A}{\cos A(\sin A-\cos A)}-\frac{\cos ^2 A}{\sin A(\sin A-\cos A)}$
$=\frac{\sin ^3 A-\cos 3 A}{\sin A \cdot \cos A(\sin A-\cos A)}$
$=\frac{1+\sin A \cdot \cos A}{\sin A \cdot \cos A}$
$=\frac{1}{\sin A \cdot \cos A}+\frac{\sin A \cdot \cos A}{\sin A \cdot \cos A}$
$= \sec A.\ce{cosec} \ A + 1$
$= \ce{RHS}$
Hence proved.
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