anonymous
  • anonymous
Trig: How does this equal one? \( tan^2 \theta - sec^2 \theta = 1 \)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I have \( tan^2 \theta - sec^2 \theta = 1 \) \( \frac{sin^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta} \) I am stuck here
anonymous
  • anonymous
I think I got it
Astrophysics
  • Astrophysics
It shouldn't

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anonymous
  • anonymous
Let me know if this is correct \( tan^2 \theta - sec^2 \theta = 1 \) \( \frac{sin^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta} \) \( \frac{1+cos^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta} \) \( \frac{cos^2 \theta}{cos^2 \theta} = 1 \)
freckles
  • freckles
do you mean -1? recall \[\sin^2(\theta)+\cos^2(\theta)=1 \] divide both sides by cos^2(theta) (this equation comes |dw:1434076836880:dw|)
anonymous
  • anonymous
Yes I mean - 1
Astrophysics
  • Astrophysics
\[\sec^2 \theta - \tan^2 \theta = 1\]
anonymous
  • anonymous
What I did is correct? Besides leaving off - ??
anonymous
  • anonymous
This is correct? \( tan^2 \theta - sec^2 \theta = -1 \) \( \frac{sin^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta} \) \( \frac{1-cos^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta} \) \( \frac{-cos^2 \theta}{cos^2 \theta} = -1 \)
Astrophysics
  • Astrophysics
\[\sec^2 \theta - \tan^2 \theta = 1 \implies \frac{ 1 }{ \cos^2 \theta } - \frac{ \sin^2 \theta }{ \cos^2 \theta } = 1\] \[\frac{ 1- \sin ^2 \theta }{ \cos^2 \theta } = 1 \implies \frac{ \cos^2 \theta }{ \cos^2 \theta } = 1 \implies 1 = 1\]
Astrophysics
  • Astrophysics
Yeah, that looks good
anonymous
  • anonymous
Ok cool. Just needed to know I was on the right path. Thanks
Astrophysics
  • Astrophysics
np :)

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