Question
Prove that $\frac{\sec A }{\tan A +\cot A }=\sin A$
✓
Answer
$\text { L.H.S }=\frac{\sec A }{\tan A +\cot A }$
$=\frac{\sec A }{\frac{\sin A }{\cos A }+\frac{\cos A }{\sin A }}$
$=\frac{\sec A }{\frac{\sin ^2 A +\cos ^2 A }{\cos A \sin A }}$
$=\frac{\sec A }{\frac{1}{\cos A \sin A }} \cdots \ldots \cdot\left[\cdot \sin ^2 A +\cos ^2 A =1\right]$
$=\sec A \cos A \sin A$
$=\frac{1}{\cos A } \times \cos A \sin A$
$=\sin A$
$=\text { R.H.S. }$
$\therefore \frac{\sec A }{\tan A +\cot A }=\sin A $
$=\frac{\sec A }{\frac{\sin A }{\cos A }+\frac{\cos A }{\sin A }}$
$=\frac{\sec A }{\frac{\sin ^2 A +\cos ^2 A }{\cos A \sin A }}$
$=\frac{\sec A }{\frac{1}{\cos A \sin A }} \cdots \ldots \cdot\left[\cdot \sin ^2 A +\cos ^2 A =1\right]$
$=\sec A \cos A \sin A$
$=\frac{1}{\cos A } \times \cos A \sin A$
$=\sin A$
$=\text { R.H.S. }$
$\therefore \frac{\sec A }{\tan A +\cot A }=\sin A $
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