anonymous one year ago Trig: How does this equal one? $$tan^2 \theta - sec^2 \theta = 1$$

1. 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

2. anonymous

I think I got it

3. Astrophysics

It shouldn't

4. anonymous

Let me know if this is correct $$tan^2 \theta - sec^2 \theta = 1$$ $$\frac{sin^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta}$$ $$\frac{1+cos^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta}$$ $$\frac{cos^2 \theta}{cos^2 \theta} = 1$$

5. 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|)

6. anonymous

Yes I mean - 1

7. Astrophysics

$\sec^2 \theta - \tan^2 \theta = 1$

8. anonymous

What I did is correct? Besides leaving off - ??

9. anonymous

This is correct? $$tan^2 \theta - sec^2 \theta = -1$$ $$\frac{sin^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta}$$ $$\frac{1-cos^2 \theta}{cos^2 \theta} - \frac{1}{cos^2 \theta}$$ $$\frac{-cos^2 \theta}{cos^2 \theta} = -1$$

10. Astrophysics

$\sec^2 \theta - \tan^2 \theta = 1 \implies \frac{ 1 }{ \cos^2 \theta } - \frac{ \sin^2 \theta }{ \cos^2 \theta } = 1$ $\frac{ 1- \sin ^2 \theta }{ \cos^2 \theta } = 1 \implies \frac{ \cos^2 \theta }{ \cos^2 \theta } = 1 \implies 1 = 1$

11. Astrophysics

Yeah, that looks good

12. anonymous

Ok cool. Just needed to know I was on the right path. Thanks

13. Astrophysics

np :)