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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Updated on: 10-Oct-2022

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