What is $sec^2theta-csc^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2015-12-06T04:54:15

$(sectheta+csctheta)(sectheta-csctheta)$

This is a difference of squares.

The general rule is that $a^2-b^2=(a+b)(a-b)$ .

Because of this:

$sec^2theta-csc^2theta=(sectheta+csctheta)(sectheta-csctheta)$

What is sec^2theta-csc^2thetasec^2theta-csc^2theta in terms of non-exponential trigonometric functions?