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anonymous

  • one year ago

Prove - tan^2(x) + sec^2(x) = 1 by working on one side to match the other using identities.

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

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

  3. Nnesha
    • one year ago
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    sec^2 theta = what ? remember the identity ?

  4. anonymous
    • one year ago
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    1/cos^2x

  5. Nnesha
    • one year ago
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    well that's reciprocal of sec but it's okay we can use that too!! tan^2 =what ?

  6. anonymous
    • one year ago
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    sin^2x/cos^2x

  7. anonymous
    • one year ago
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    or 1/cot^2x

  8. Nnesha
    • one year ago
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    yes right so replace tan and sec with that \[\huge\rm -\frac{ \sin^2x }{ \cos^2x } +\frac{ 1 }{ \cos^2x}\] find the common denominator

  9. anonymous
    • one year ago
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    cos^2x?

  10. Nnesha
    • one year ago
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    ohh well not gonna work should use the identity i guess

  11. Jhannybean
    • one year ago
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    You could also use the fact that \(\sf sec^2(\theta) = tan^2(\theta) +1\) and then substitute this in place of \(\sf \sec^2(\theta)\)

  12. anonymous
    • one year ago
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    thats true, thanks

  13. Jhannybean
    • one year ago
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    \[\sf -tan^2(\theta)+\sec^2(\theta) = 1\]\[\sf -\tan^2(\theta) +\color{red}{\tan^2(\theta) +1}=1\]

  14. anonymous
    • one year ago
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    Wow... The one identity I didn't think of solved it so easily. Thank you!

  15. Nnesha
    • one year ago
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    \[\huge\rm \frac{ -\sin^2x +1}{ \cos^2 }\] use the special identity sin^2x+cos^2x =1 solve for cos^2

  16. Jhannybean
    • one year ago
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    No problem :)

  17. Nnesha
    • one year ago
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    here you can copy these identities http://www.math.com/tables/trig/identities.htm you weren't familiar with this so that's why i thought better to write interms of sin and cos

  18. Jhannybean
    • one year ago
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    That's a good way to approach it too. @Nnesha :)

  19. anonymous
    • one year ago
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    Thanks that will help too @Nnesha

  20. Nnesha
    • one year ago
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    yw :=)

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