Question:
Prove the following trigonometric identities.
$\sec A(1-\sin A)(\sec A+\tan A)=1$
Solution:
We have to prove $\sec A(1-\sin A)(\sec A+\tan A)=1$
We know that, $\sec ^{2} A-\tan ^{2} A=1$
So,
$\sec A(1-\sin A)(\sec A+\tan A)=\{\sec A(1-\sin A)\}(\sec A+\tan A)$
$=(\sec A-\sec A \sin A)(\sec A+\tan A)$
$=\left(\sec A-\frac{1}{\cos A} \sin A\right)(\sec A+\tan A)$
$=\left(\sec A-\frac{\sin A}{\cos A}\right)(\sec A+\tan A)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^{2} A-\tan ^{2} A$
$=1$
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