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Prove the following trigonometric identities:
\( \sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1 \)
To do:
We have to prove that \( \sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1 \).
Solution:
We know that,
$\sec ^{2} \theta-\tan^2 \theta=1$.......(i)
$(a+b)^3=a^3+b^3+3ab(a+b)$.........(ii)
Therefore,
$\sec ^{6} \theta=(\sec ^2 \theta)^3$
$=(1+\tan^2 \theta)^3$ [From (i)]
$=1^3+(\tan^2 \theta)^3+3(1)(\tan^2 \theta)(1+\tan^2 \theta)$ [From (ii)]
$=1+\tan^6 \theta+3\tan^2 \theta(1+\tan^2 \theta)$
$=1+\tan^6 \theta+3\tan^2 \theta\sec^2 \theta$ [From (i)]
$=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1$
Hence proved.       
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