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Integration by Substitution

Prove: √ 1+s...

Question

Prove:
$\sqrt{\frac{1 + s i n A}{1 - s i n A}} = s e c A + t a n A$

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Solution

$\sqrt{\frac{1 + sin A}{1 - sin A}} = \sqrt{\frac{(1 + sin A)^{2}}{(1 - sin A) (1 + sin A)}} = \sqrt{\frac{(1 + sin A)^{2}}{{cos}^{2} A}} = \frac{1 + sin A}{cos A} = \frac{1}{cos A} + \frac{sin A}{cos A} = sec A + tan A$

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Similar questions

\frac{1 - s i n A}{1 + s i n A} = (s e c A - t a n A)^{2}

{(s e c A - t a n A)}^{2} = \frac{1 - s i n A}{1 + s i n A}

secA - t a n A = \frac{cos A}{1 + sin A}

Q. Prove that tanA+secA-1/tanA-secA+1=1+sinA/cosA

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

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