Question
Prove.
$\frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec}^2 A}=\tan A$
$\frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec}^2 A}=\tan A$
✓
Answer
$\text { LHS }=\frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec} A} $
$\frac{\sec ^2 A \cot A}{\cos e c^2 A}\left(\because \sec ^2 A =1+\tan ^2 A \right) $
$\frac{\frac{1}{\cos ^2 A} \times \frac{\cos A}{\sin A}}{\frac{1}{\sin ^2 A}}=\frac{\frac{1}{\cos A \sin A}}{\frac{1}{\sin ^2 A}}$
$=\frac{\sin A}{\cos A}=\tan A=\text { RHS }$
$\frac{\sec ^2 A \cot A}{\cos e c^2 A}\left(\because \sec ^2 A =1+\tan ^2 A \right) $
$\frac{\frac{1}{\cos ^2 A} \times \frac{\cos A}{\sin A}}{\frac{1}{\sin ^2 A}}=\frac{\frac{1}{\cos A \sin A}}{\frac{1}{\sin ^2 A}}$
$=\frac{\sin A}{\cos A}=\tan A=\text { RHS }$
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