Question
Prove that: $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$
✓
Answer
$=\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$
$=\sqrt{\frac{(1+\sin A)^2}{1-\sin ^2 A}}=\sqrt{\frac{(1+\sin A)^2}{\cos ^2 A}}$
$=\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}$
$=\sec A +\tan A$
$=\text { RHS }$
$=\sqrt{\frac{(1+\sin A)^2}{1-\sin ^2 A}}=\sqrt{\frac{(1+\sin A)^2}{\cos ^2 A}}$
$=\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}$
$=\sec A +\tan A$
$=\text { RHS }$
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