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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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