Question
Prove the following trigonometric identities.
$\sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1$
$\sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1$
✓
Answer
We know that $\sec^2\theta-\tan^2\theta=1$
Cubing on both sides
$\text{L.H.S}=(\sec^2\theta-\tan^2\theta)^3=1^3$
$\sec^6\theta-\tan^6\theta-3\sec^2\theta\tan^2\theta(\sec^2\theta-\tan^2\theta)=1$
$\tan^2\theta \big[\because (\text{a}-\text{b})^2=\text{a}^3-\text{b}^3-3\text{ab}(\text{a}-\text{b})\big]$
$\Rightarrow\ \sec^6\theta-\tan^6\theta-3\sec^2\theta\tan^2\theta=1$
$\Rightarrow\ \sec^6\theta=\tan^6\theta+1+3\tan^2\theta\sec^2\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
Cubing on both sides
$\text{L.H.S}=(\sec^2\theta-\tan^2\theta)^3=1^3$
$\sec^6\theta-\tan^6\theta-3\sec^2\theta\tan^2\theta(\sec^2\theta-\tan^2\theta)=1$
$\tan^2\theta \big[\because (\text{a}-\text{b})^2=\text{a}^3-\text{b}^3-3\text{ab}(\text{a}-\text{b})\big]$
$\Rightarrow\ \sec^6\theta-\tan^6\theta-3\sec^2\theta\tan^2\theta=1$
$\Rightarrow\ \sec^6\theta=\tan^6\theta+1+3\tan^2\theta\sec^2\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
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