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

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LynFran
 one year ago
Best ResponseYou've already chosen the best response.3tan2x+sec2x \[\frac{ \sin2x }{ \cos2x }+\frac{ 1 }{ \cos2x }\]\[\frac{ \sin2x+1 }{ \cos2x }\]\[1\sin2x=\cos2x\]so\[\frac{ \cos2x }{ \cos2x }=1\]..proven...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you sooo much @LynFran

LynFran
 one year ago
Best ResponseYou've already chosen the best response.3ur welcome..please medal..thanks

LynFran
 one year ago
Best ResponseYou've already chosen the best response.3@DecentNabeel its the Pythagorean identity\[cosx+sinx=1\]so \[\cos2x+\sin2x=1 \] so \[\cos2x=1\sin2x\]

LynFran
 one year ago
Best ResponseYou've already chosen the best response.3@freckles please tell @DecentNabeel im right

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I'm gonna have to agree with @DecentNabeel here but I do think he meant to write \[\tan^2(x)+\sec^2(x)=1 \text{ instead of } \tan(2x)+\sec(2x)=1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that last I was thinking about the OP when I said he

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan ^2x+\sec ^2x=1\]

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Is that^^^what you are trying to prove?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sin(x)+\cos(x) \neq 1 \text{ for all } x \\ \text{ the pythagorean identity is } \\ \sin^2(x)+\cos^2(x)=1\] dw:1435970814153:dw we see here the following: \[\cos(\theta)=\frac{x}{r} \implies r \cos(\theta)= x \\ \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)= y \\ \text{ now by Pythagorean theorem we have } \\ x^2+y^2=r^2 \\(r \cos(\theta))^2+(r \sin(\theta))^2=r^2 \\ r^2 \cos^2(\theta)+r^2 \sin^2(\theta)=r^2 \\ \text{ divide both sides by } r^2 \\ \cos^2(\theta)+\sin^2(\theta)=1 \]

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0I see that the asker has disappeared.
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