Prove the following trigonometric identities:$ \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,

$\sin ^{2} A+\cos^2 A=1$.......(i)

$\sec A=\frac{1}{\cos A}$......(ii)

$\tan A=\frac{\sin A}{\cos A}$.......(iii)

Therefore,

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$            (Multiplying and dividing by $1+\sin A$)

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

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

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

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

$=\sec A+\tan A$

Hence proved.  

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Updated on: 10-Oct-2022

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