Question
Prove the following trigonometric identities.
$\sec^4\text{A}(1-\sin^4\text{A})-2\tan^2\text{A}=1$
$\sec^4\text{A}(1-\sin^4\text{A})-2\tan^2\text{A}=1$
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Answer
$\text{L.H.S}=\sec^4\text{A}(1-\sin^4\text{A})-2\tan^2\text{A}$
$=\sec^4\text{A}-\sec^4\text{A}\times\sin^4\text{A}-2\tan^2\text{A}$
$=\sec^4\text{A}-\frac{1}{\cos^4\text{A}}\times\sin^4\text{A}-2\tan^2\text{A}$
$=\sec^4\text{A}-\tan^4\text{A}-2\tan^2\text{A}$
$=(\sec^2\text{A})^2-\tan^4\text{A}-2\tan^2\text{A}$
$=(1+\tan^2\text{A})^2-\tan^4\text{A}-2\tan^2\text{A}\ \big[\because \sec^2\text{A}-\tan^2\text{A}=1\big]$
$=1+\tan^4\text{A}+2\tan^2\text{A}-\tan^4\text{A}-2\tan^2\text{A}$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$=\sec^4\text{A}-\sec^4\text{A}\times\sin^4\text{A}-2\tan^2\text{A}$
$=\sec^4\text{A}-\frac{1}{\cos^4\text{A}}\times\sin^4\text{A}-2\tan^2\text{A}$
$=\sec^4\text{A}-\tan^4\text{A}-2\tan^2\text{A}$
$=(\sec^2\text{A})^2-\tan^4\text{A}-2\tan^2\text{A}$
$=(1+\tan^2\text{A})^2-\tan^4\text{A}-2\tan^2\text{A}\ \big[\because \sec^2\text{A}-\tan^2\text{A}=1\big]$
$=1+\tan^4\text{A}+2\tan^2\text{A}-\tan^4\text{A}-2\tan^2\text{A}$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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