## anonymous one year ago how do i prove this????? pls help - tan2x + sec2x = 1

1. LynFran

-tan2x+sec2x $\frac{ -\sin2x }{ \cos2x }+\frac{ 1 }{ \cos2x }$$\frac{ -\sin2x+1 }{ \cos2x }$$1-\sin2x=\cos2x$so$\frac{ \cos2x }{ \cos2x }=1$..proven...

2. anonymous

thank you sooo much @LynFran

3. LynFran

4. LynFran

@DecentNabeel its the Pythagorean identity$cosx+sinx=1$so $\cos2x+\sin2x=1$ so $\cos2x=1-\sin2x$

5. LynFran

@freckles please tell @DecentNabeel im right

6. freckles

I'm gonna have to agree with @DecentNabeel here but I do think he meant to write $-\tan^2(x)+\sec^2(x)=1 \text{ instead of } -\tan(2x)+\sec(2x)=1$

7. freckles

that last I was thinking about the OP when I said he

8. Mertsj

$-\tan ^2x+\sec ^2x=1$

9. Mertsj

Is that^^^what you are trying to prove?

10. freckles

$\sin(x)+\cos(x) \neq 1 \text{ for all } x \\ \text{ the pythagorean identity is } \\ \sin^2(x)+\cos^2(x)=1$ |dw:1435970814153:dw| we see here the following: $\cos(\theta)=\frac{x}{r} \implies r \cos(\theta)= x \\ \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)= y \\ \text{ now by Pythagorean theorem we have } \\ x^2+y^2=r^2 \\(r \cos(\theta))^2+(r \sin(\theta))^2=r^2 \\ r^2 \cos^2(\theta)+r^2 \sin^2(\theta)=r^2 \\ \text{ divide both sides by } r^2 \\ \cos^2(\theta)+\sin^2(\theta)=1$

11. Mertsj

I see that the asker has disappeared.