MHB Can you prove the following two difficult trigonometric identities?

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The discussion focuses on proving two challenging trigonometric identities involving secant, tangent, sine, and cosine functions. Participants are encouraged to use the available LaTeX compiler for clear expression of mathematical equations. A free math tutoring video is suggested for those needing assistance with the proof methods. The conversation highlights the importance of community support in tackling complex mathematical problems. Overall, the forum serves as a resource for learning and collaboration in mathematics.
DrLiangMath
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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:

 
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@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.
[math]sec^6(x) - tan^6(x) = 1 + 3 ~tan^2(x) ~ sec^2(x)[/math]

[math]sin^2(x) ~ tan(x) + cos^2(x) ~ cot(x) + 2 ~ sin(x) ~ cos(x) = tan(x) + cot(x)[/math]

-Dan
 
Hello Dan,

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

Derek
 
This is useful information
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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