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anonymous

  • one year ago

how do i prove this????? pls help - tan2x + sec2x = 1

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  1. LynFran
    • one year ago
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    -tan2x+sec2x \[\frac{ -\sin2x }{ \cos2x }+\frac{ 1 }{ \cos2x }\]\[\frac{ -\sin2x+1 }{ \cos2x }\]\[1-\sin2x=\cos2x\]so\[\frac{ \cos2x }{ \cos2x }=1\]..proven...

  2. anonymous
    • one year ago
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    thank you sooo much @LynFran

  3. LynFran
    • one year ago
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    ur welcome..please medal..thanks

  4. LynFran
    • one year ago
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    @DecentNabeel its the Pythagorean identity\[cosx+sinx=1\]so \[\cos2x+\sin2x=1 \] so \[\cos2x=1-\sin2x\]

  5. LynFran
    • one year ago
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    @freckles please tell @DecentNabeel im right

  6. freckles
    • one year ago
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    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
    • one year ago
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    that last I was thinking about the OP when I said he

  8. Mertsj
    • one year ago
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    \[-\tan ^2x+\sec ^2x=1\]

  9. Mertsj
    • one year ago
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    Is that^^^what you are trying to prove?

  10. freckles
    • one year ago
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    \[\sin(x)+\cos(x) \neq 1 \text{ for all } x \\ \text{ the pythagorean identity is } \\ \sin^2(x)+\cos^2(x)=1\] |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
    • one year ago
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    I see that the asker has disappeared.

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