Prove that $ \sqrt{\frac{1+\sin A}{1-\sin A }}=\sec A+\tan A $.

To do:

We have to prove that \( \sqrt{\frac{1+\sin A}{1-\sin A }}=\sec A+\tan A \).

Solution:

We know that,

$\frac{1}{cos\ A}=sec\ A$

$\frac{sin\ A}{cos\ A}=tan\ A$
$sin^2\ A+cos^2\ A=1$

LHS

$\sqrt{\frac{1+\sin A}{1-\sin A }}=(\sqrt{\frac{1+\sin A}{1-\sin A }})\times\sqrt{\frac{1+\sin A}{1+\sin A }}$     (Rationalising the denominator)

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

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

$=\frac{1+\sin A}{cos\ A}$

$=\frac{1}{cos\ A}+\frac{sin\ A}{cos\ A}$

$=sec\ A+tan\ A$

Hence proved.

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Updated on: 10-Oct-2022

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