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Prove the following trigonometric identities:$ \frac{1}{1+\sin A}+\frac{1}{1-\sin A}=2 \sec ^{2} A $
To do:
We have to prove that \( \frac{1}{1+\sin A}+\frac{1}{1-\sin A}=2 \sec ^{2} A \).
Solution:
We know that,
$\sec A=\frac{1}{\cos A}$.....(i)
$\cos ^{2} A+\sin^2 A=1$.......(ii)
Therefore,
$\frac{1}{1+\sin A}+\frac{1}{1-\sin A}=\frac{1-\sin A+1+\sin A}{(1-\sin A)(1+\sin A)}$
$=\frac{2}{1-\sin^2 A}$
$=\frac{2}{\cos^2 A}$ [From (ii)]
$=2 \sec^2 A$ [From (i)]
Hence proved.   
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