Can you prove the following two difficult trigonometric identities?

[DrLiangMath]

Can you prove the following?

[sec(x)]^6 - [tan(x)]^6 = 1 + 3*[tan(x)]^2*[sec(x)]^2

[sin(x)]^2*tan(x) + [cos(x)]^2*cot(x) + 2*sin(x)*cos(x) = tan(x) + cot(x)

If not, the following free math tutoring video shows you the method:

 

[topsquark]

@Dr. Liang: We have a LaTeX compiler here that you can use to type out the equations. If you don't know LaTeX reply to this to see how the coding works. The basics are simple and we have a Forum to show you how to do more complicated work.
\(\displaystyle sec^6(x) - tan^6(x) = 1 + 3 ~tan^2(x) ~ sec^2(x)\)

\(\displaystyle sin^2(x) ~ tan(x) + cos^2(x) ~ cot(x) + 2 ~ sin(x) ~ cos(x) = tan(x) + cot(x)\)

-Dan

 

[DrLiangMath]

Hello Dan,

Thank you so much for your kindness and help! I didn't notice a LaTeX compiler is available in the forum.

Derek

 

[Exbar15]

This is useful information