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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Updated on: 10-Oct-2022

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