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anonymous

  • one year ago

prove this trigonometric equation; - tan^2x + sec^2x = 1

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

  2. UsukiDoll
    • one year ago
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    hmm that's one of the identities only rearranged...

  3. UsukiDoll
    • one year ago
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    there is a trig identity which is \[\tan^2x+1 =\sec^2x \]

  4. anonymous
    • one year ago
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    yes, that is correct (the first response you made)

  5. UsukiDoll
    • one year ago
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    @lxoser subtract \[\tan^2x \] on both sides for \[\tan^2x+1 =\sec^2x \]

  6. anonymous
    • one year ago
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    1 = sec^2x - tan^2x ?

  7. UsukiDoll
    • one year ago
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    yeah.. now if we rearrange this \[1=\sec^2x-\tan^2x \] to \[1= -\tan^2x+\sec^2x\]

  8. AakashSudhakar
    • one year ago
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    Hm, if I recall correctly, this identity can be derived from the following expression: \[\sin^2x + \cos^2x = 1\]It's actually quite simple. Simply divide the entire equation by [cos(x)]^2. Then simply each individual term!

  9. UsukiDoll
    • one year ago
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    I think there's already an identity which is \[\tan^2x+1 =\sec^2x \] then just rearrange the terms.

  10. UsukiDoll
    • one year ago
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    oh I see where you're going on this... xD

  11. AakashSudhakar
    • one year ago
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    Oh, I wasn't sure whether or not he wanted that identity derived as well. Either way, rearranging the terms becomes simple once you know the direction to go in!

  12. anonymous
    • one year ago
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    im confused now :/

  13. UsukiDoll
    • one year ago
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    there are 3 trig identities.. two of them were posted here but I think using one of the identities would make it a bit easier

  14. UsukiDoll
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @UsukiDoll there is a trig identity which is \[\tan^2x+1 =\sec^2x \] \(\color{blue}{\text{End of Quote}}\)

  15. AakashSudhakar
    • one year ago
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    @lxoser, do you know what the question exactly asks you? Does it simply ask to derive the expression you have from any trig identity, or does it want you to derive it from a specific identity? In any case, both my answer and @UsukiDoll's answers are correct, just using/manipulating different trig identities.

  16. UsukiDoll
    • one year ago
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  17. anonymous
    • one year ago
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    the full question is ; verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation

  18. UsukiDoll
    • one year ago
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    it's one of those trig proofs... so start from the left to achieve the right the thing is that whatever was presented to us is similar to a trig identity, so we can use that particular trig identity, manipulate it a bit, and then we got the right side

  19. AakashSudhakar
    • one year ago
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    Unfortunately, I'm not quite sure what that means. Does it want both sides of the equation to be equal?

  20. anonymous
    • one year ago
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    okay so what i understand is - tan^2x + sec^2x = 1 is just the reverse of the trigonometric identity tan^2x + 1 = sec^2x & yes pretty much.

  21. UsukiDoll
    • one year ago
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    \[-\tan^2x+\sec^2x=1 \] using the identity \[\tan^2x+1 =\sec^2x \] subtract \[\tan^2x \] from both sides \[1 =\sec^2x-tan^2x \] rearrange \[1 =-tan^2x+sec^2x \]

  22. anonymous
    • one year ago
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    is that the final answer?

  23. UsukiDoll
    • one year ago
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    if we substitute either way, it's equal like.. \[1 =-\tan^2x+\sec^2x \] As in 1 is equal to that ^ then sub it to the original question -\tan^2x+\sec^2x= (-\tan^2x+\sec^2x) both equal similarly 1 =1

  24. anonymous
    • one year ago
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    okay, i see now.

  25. UsukiDoll
    • one year ago
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    ugh my latex broke

  26. UsukiDoll
    • one year ago
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    but we got them to equal.. that's the main thing

  27. anonymous
    • one year ago
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    i understand now, thank you sooo much!!

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