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

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 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^2xtan^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.

Aand b is what i think

6/9 i need more characters in my answer