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

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Answer 1 · 2015-12-06T04:54:15

$(sec θ + csc θ) (sec θ - csc θ)$ $(sectheta+csctheta)(sectheta-csctheta)$

This is a difference of squares.

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

Because of this:

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

What is sec2θ−csc2θsec^2theta-csc^2theta in terms of non-exponential trigonometric functions?