Question
Prove that $\frac{1+\sec A }{\sec A }=\frac{\sin ^2 A }{1-\cos A }$
✓
Answer
$ \text { L.H.S }=\frac{1+\sec A}{\sec A}$
$ =\frac{1}{\sec A}+\frac{\sec A}{\sec A}$
$ =\cos A+1$
$ =(1+\cos A) \times \frac{1-\cos A}{1-\cos A}$
$ =\frac{1-\cos ^2 A}{1-\cos A}$
$=\frac{\sin ^2 A}{1-\cos A} ........ \left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =\text { R.H.S }$
$ \therefore \frac{1+\sec A}{\sec A}=\frac{\sin ^2 A}{1-\cos A}$
$ =\frac{1}{\sec A}+\frac{\sec A}{\sec A}$
$ =\cos A+1$
$ =(1+\cos A) \times \frac{1-\cos A}{1-\cos A}$
$ =\frac{1-\cos ^2 A}{1-\cos A}$
$=\frac{\sin ^2 A}{1-\cos A} ........ \left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =\text { R.H.S }$
$ \therefore \frac{1+\sec A}{\sec A}=\frac{\sin ^2 A}{1-\cos A}$
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