Trig Identity: sec^6x - tan^6x = 1 + 3tan^2sec^2x

[Spoon-]

I can't solve this identity help please!

sec[sup:3suhmy47]6[/sup:3suhmy47]x - tan[sup:3suhmy47]6[/sup:3suhmy47]x = 1 + 3tan[sup:3suhmy47]2[/sup:3suhmy47]sec[sup:3suhmy47]2[/sup:3suhmy47]x

 

[wjm11]

Re: Trig Identity

I think the following hints may prove helpful:
Difference of two squares factoring formula: a^2 - b^2 = (a+b)(a-b)
a^6 - b^6 = (a^3)^2 - (b^3)^2 = (a^3 - b^3)(a^3 + b^3)
or
a^6 - b^6 = (a^2)^3 - (b^2)^3
There are factoring formulas for both the sum and difference of two cubes as well:
http://www.math.unt.edu/mathlab/emathla ... ocubes.htm
Also recall that
sec^2(x) - tan^2(x) = 1

Hope that helps.

 

[Deleted member 4993]

Re: Trig Identity

Also use:

\(\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)