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8. Introduction to Trigonometry
medium
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\sqrt{\frac{1+\sin A }{1-\sin A }}=\sec A +\tan A$
Option A
Option B
Option C
Option D
Solution
$\sqrt{\frac{1+\sin A }{1-\sin A }}=\sec A +\tan A$
$L.H.S.=\sqrt{\frac{1+\sin A }{1-\sin A }}$
$=\sqrt{\frac{(1+\sin A )(1+\sin A )}{(1-\sin A )(1+\sin A )}}$
$=\frac{(1+\sin A )}{\sqrt{1-\sin ^{2} A }}=\frac{1+\sin A }{\sqrt{\cos ^{2} A }}$
$=\frac{1+\sin A }{\cos A } \quad=\sec A +\tan A$
$= R . H.S.$
Standard 10
Mathematics
