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 $sec\ A\ (1 − \sin A)(sec\ A + \tan A) = 1$
We know that $sec^2\ A − \tan^2 A − 1$
So$,$
$sec\ A(1 − \sin A)(sec\ A + \tan A) = {sec \ A(1 − \sin A)}(sec A + \tan A)$
$= (sec\ A − sec\ A \sin A)(sec\ A + \tan A)$
$=\left(\sec\ A-\frac{1}{\cos A} \sin A\right)(\sec\ A+\tan A) \ldots\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$=\left(\sec\ A-\frac{\sin A}{\cos A}\right)(\sec\ A+\tan A) \ldots\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^2 A-\tan ^2 A$
$=1=\text { R.H.S. }\left(\because \sec ^2 \theta=1 \tan ^2 \theta\right)$
We know that $sec^2\ A − \tan^2 A − 1$
So$,$
$sec\ A(1 − \sin A)(sec\ A + \tan A) = {sec \ A(1 − \sin A)}(sec A + \tan A)$
$= (sec\ A − sec\ A \sin A)(sec\ A + \tan A)$
$=\left(\sec\ A-\frac{1}{\cos A} \sin A\right)(\sec\ A+\tan A) \ldots\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$=\left(\sec\ A-\frac{\sin A}{\cos A}\right)(\sec\ A+\tan A) \ldots\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^2 A-\tan ^2 A$
$=1=\text { R.H.S. }\left(\because \sec ^2 \theta=1 \tan ^2 \theta\right)$
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