Math (Double checking trig identity)

Since sec^2θ - 1 = tan^2θ

I know this is trivial, but I want to make sure I'm doing this right before I apply it to the integral I'm trying to solve...

If I have some constant a,

(a^2secθ)^2 - (a^2)^2

If I wanted to change this to tan would it be:

a^4tan^4θ?

Any help is greatly appreciated!

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asked by Ray
  1. not at all.

    (a^2secθ)^2 - (a^2)^2
    = a^4 (sec^4θ - 1)
    = a^4 (sec^2θ-1)(sec^2θ+1)
    = a^4 tan^2θ (sec^2θ+1)

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    posted by Steve
  2. Nope. my bad. Still you were also wrong

    (a^2secθ)^2 - (a^2)^2
    = a^4 (sec^2θ-1)
    = a^4 tan^2θ

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    posted by Steve
  3. So tan^2θ is where I messed up, thanks Steve!

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    posted by Ray

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