# 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.   $(vi)\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A$

M manish

We need to prove -
$\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A$
Taking LHS;
By rationalising the denominator, we get;

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

Hence proved.

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