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