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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.