Question
Prove the following trigonometric identities.
$\frac{1}{\sec\text{A}+\tan\text{A}}-\frac{1}{\cos\text{A}}=\frac{1}{\cos\text{A}}-\frac{1}{\sec\text{A}-\tan\text{A}}$

Answer

$\text{L.H.S}: \sec\text{A}-\tan\text{A}\Big[\therefore\ \frac{1}{\sec\text{A}+\tan\text{A}}=\sec\text{A}-\tan\text{A}\Big]$
$=-\tan\text{A}$
$\text{R.H.S}\frac{1}{\cos\text{A}}-\frac{1}{\sec\text{A}-\tan\text{A}}$
$\sec\text{A}-(\sec\text{A}+\tan\text{A})$
$\Big[\therefore\ \frac{1}{\sec\text{A}-\tan\text{A}}=\sec\text{A}+\tan\text{A}\Big]$
$=-\tan\text{A}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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