Prove the following trigonometric identities:$ \frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\cos ^{2} A}{(1+\sin A)^{2}} $

To do:

We have to prove that \( \frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\cos ^{2} A}{(1+\sin A)^{2}} \).

Solution:

We know that,

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

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

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

Therefore,

$\frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\frac{1}{\cos A}-\frac{\sin A}{\cos A}}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}$

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

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

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

Multiplying and dividing by ($1+\sin A$), we get,

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

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

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

Hence proved.       

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Updated on: 10-Oct-2022

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