$\sqrt{\frac{1 + sin A}{1 - sinA}} = secA + tan A$

True

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False

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Solution

The correct option is A True
$\frac{1 + sin A}{1 - sin A} = \frac{(1 + sin A) (1 + sin A)}{(1 - sin A) (1 + sin A)} = \frac{(1 + sin A)^{2}}{1 - {sin}^{2} A} = \frac{(1 + sin A)^{2}}{{cos}^{2} A} = (\frac{1 + sin A}{cos A})^{2} = (sec A + tan A)^{2}$

So $\sqrt{\frac{1 + sin A}{1 - sin A}} = sec A + tan A$

So the relation is $True$

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