Question

Prove identity:
Sec^6x-tan^6x= 1+3tan^2xsec^2x

Answer 1

This time, we start from the left side.
\(\sec^6x-\tan^6x\)
\(=(\sec^2x)^3-(\tan^2x)^3\)

Then, use the identity \(a^3-b^3 = (a-b)(a^2+ab+b^2)\), we'll get
\((\sec^2x-\tan^2x)(\sec^4x+\sec^2x\tan^2x+\tan^4x)\)

Since \(\tan^2x+1=\sec^2x\), we have \((\sec^2x-\tan^2x) = 1\). So,
\((\sec^2x-\tan^2x)(\sec^4x+\sec^2x\tan^2x+\tan^4x)\)
\(=\sec^4x+\sec^2x\tan^2x+\tan^4x\)

Then consider \(\sec^4x\),
\(\sec^4x = \sec^2x (\sec^2x) = \sec^2x(\tan^2x+1) = \sec^2x\tan^2x+\sec^2x\)

Consider \(\tan^4x\),
\(\tan^4x = \tan^2x (\tan^2x) = \tan^2x(\sec^2x+1) = \sec^2x\tan^2x-\tan^2x\)

Substitute the two back to \(\sec^4x+\sec^2x\tan^2x+\tan^4x\) , and simplify it
With the help of the identity \((\sec^2x-\tan^2x) = 1\), you should be able to get the right side.