Question
Prove the following trigonometric identities.
$sec A (1 − sin A) (sec A + tan A) = 1$
$sec A (1 − sin A) (sec A + tan A) = 1$
✓
Answer
We have to prove $\sec \mathrm{A}(1-\sin \mathrm{A})(\sec \mathrm{A}+\tan \mathrm{A})=1$
We know that $\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}-1$
So,
$\sec A(1-\sin A)(\sec A+\tan A)=\{\sec A(1-\sin A)\}(\sec A+\tan A)$
$=(\sec A-\sec A \sin A)(\sec A+\tan A)$
$=\left(\sec A-\frac{1}{\cos A} \sin A\right)(\sec A+\tan A) \ldots\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$=\left(\sec A-\frac{\sin A}{\cos A}\right)(\sec A+\tan A) \ldots\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^2 A-\tan ^2 A$
$=1=\text { R.H.S. }\left(\because \sec ^2 \theta=1 \tan ^2 \theta\right)$
We know that $\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}-1$
So,
$\sec A(1-\sin A)(\sec A+\tan A)=\{\sec A(1-\sin A)\}(\sec A+\tan A)$
$=(\sec A-\sec A \sin A)(\sec A+\tan A)$
$=\left(\sec A-\frac{1}{\cos A} \sin A\right)(\sec A+\tan A) \ldots\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$=\left(\sec A-\frac{\sin A}{\cos A}\right)(\sec A+\tan A) \ldots\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^2 A-\tan ^2 A$
$=1=\text { R.H.S. }\left(\because \sec ^2 \theta=1 \tan ^2 \theta\right)$
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