Question
Prove the following trigonometric identities.
$(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)=2\tan\text{A}$
$(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)=2\tan\text{A}$
✓
Answer
We have to prove $(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)=2\tan\text{A}$
We know that, $\sec^2\text{A}-\tan^2\text{A}=1$
So, we have
$\text{L.H.S}=(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)$
$=\{\sec\text{A}+\tan\text{A}\}\{\sec\text{A}-(\tan\text{A}-1)\}$
$=\sec^2\text{A}-(\tan\text{A}-1)^2$
$=\sec^2\text{A}(\tan^2\text{A}-2\tan\text{A}+1)$
$=(\sec^2\text{A}-\tan^2\text{A})+2\tan\text{A}-1$
So, we have
$(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)$
$=1 + 2\tan\text{A}-1$
$= 2\tan\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
We know that, $\sec^2\text{A}-\tan^2\text{A}=1$
So, we have
$\text{L.H.S}=(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)$
$=\{\sec\text{A}+\tan\text{A}\}\{\sec\text{A}-(\tan\text{A}-1)\}$
$=\sec^2\text{A}-(\tan\text{A}-1)^2$
$=\sec^2\text{A}(\tan^2\text{A}-2\tan\text{A}+1)$
$=(\sec^2\text{A}-\tan^2\text{A})+2\tan\text{A}-1$
So, we have
$(\sec\text{A}+\tan\text{A}-1)(\sec\text{A}-\tan\text{A}+1)$
$=1 + 2\tan\text{A}-1$
$= 2\tan\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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