MHB Prove the Trigonometric Identity: $\sec^4 x - \tan^4 x = \sec^2 x + \tan^2 x$

[Jameson]

Back to trigonometry folks! Show that $\sec^4 x - \tan^4 x = \sec^2 x + \tan^2 x$.
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[Jameson]

Lots of participants this week!

Congratulations to the following members for their correct solutions:

1) MarkFL
2) Sudharaka
3) anemone
4) BAdhi
5) Petrus
6) kaliprasad
7) agentmulder
8) Reckoner

Solution (from Reckoner):

Spoiler

We will make use of the identity \[\sec^2 x - \tan^2 x = 1.\] This identity follows from the familiar Pythagorean identity, \[\sin^2 x + \cos^2 x = 1,\] if you divide both sides by $\cos^2 x$ and rearrange the terms.

By factoring the original left-hand side, we have
\begin{align*}
\sec^4 x - \tan^4 x &= \left(\sec^2 x + \tan^2 x\right)\left(\sec^2 x - \tan^2 x\right)\\
&= \left(\sec^2 x + \tan^2 x\right)(1)\\
&= \left(\sec^2 x + \tan^2 x\right).
\end{align*}

Solution (from BAdhi):

Spoiler

$$\begin{align*}
\text{L.H.S.}&= \sec^4x-\tan^4x\\
&=\left[\sec^2x\right]^2-\tan^4x\\
&=\left[1+\tan^2x\right]^2-\tan^4x\\
&=1+2\tan^2x+\tan^4x-\tan^4x\\
&=1+2\tan^2x\\
&=(\sec^2x-\tan^2x)+2\tan^2x\\
&=\sec^2x+\tan^2x
&=\text{R.H.S.}
\end{align*}$$