anonymous
  • anonymous
how do i prove this????? pls help - tan2x + sec2x = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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LynFran
  • LynFran
-tan2x+sec2x \[\frac{ -\sin2x }{ \cos2x }+\frac{ 1 }{ \cos2x }\]\[\frac{ -\sin2x+1 }{ \cos2x }\]\[1-\sin2x=\cos2x\]so\[\frac{ \cos2x }{ \cos2x }=1\]..proven...
anonymous
  • anonymous
thank you sooo much @LynFran
LynFran
  • LynFran
ur welcome..please medal..thanks

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LynFran
  • LynFran
@DecentNabeel its the Pythagorean identity\[cosx+sinx=1\]so \[\cos2x+\sin2x=1 \] so \[\cos2x=1-\sin2x\]
LynFran
  • LynFran
@freckles please tell @DecentNabeel im right
freckles
  • freckles
I'm gonna have to agree with @DecentNabeel here but I do think he meant to write \[-\tan^2(x)+\sec^2(x)=1 \text{ instead of } -\tan(2x)+\sec(2x)=1\]
freckles
  • freckles
that last I was thinking about the OP when I said he
Mertsj
  • Mertsj
\[-\tan ^2x+\sec ^2x=1\]
Mertsj
  • Mertsj
Is that^^^what you are trying to prove?
freckles
  • 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: \[\cos(\theta)=\frac{x}{r} \implies r \cos(\theta)= x \\ \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)= y \\ \text{ now by Pythagorean theorem we have } \\ x^2+y^2=r^2 \\(r \cos(\theta))^2+(r \sin(\theta))^2=r^2 \\ r^2 \cos^2(\theta)+r^2 \sin^2(\theta)=r^2 \\ \text{ divide both sides by } r^2 \\ \cos^2(\theta)+\sin^2(\theta)=1 \]
Mertsj
  • Mertsj
I see that the asker has disappeared.

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