anonymous
  • anonymous
prove this trigonometric equation; - tan^2x + sec^2x = 1
Mathematics
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SOLVED
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katieb
  • katieb
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UsukiDoll
  • UsukiDoll
\[-\tan^2x+\sec^2x=1\] this?
UsukiDoll
  • UsukiDoll
hmm that's one of the identities only rearranged...
UsukiDoll
  • UsukiDoll
there is a trig identity which is \[\tan^2x+1 =\sec^2x \]

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anonymous
  • anonymous
yes, that is correct (the first response you made)
UsukiDoll
  • UsukiDoll
@lxoser subtract \[\tan^2x \] on both sides for \[\tan^2x+1 =\sec^2x \]
anonymous
  • anonymous
1 = sec^2x - tan^2x ?
UsukiDoll
  • UsukiDoll
yeah.. now if we rearrange this \[1=\sec^2x-\tan^2x \] to \[1= -\tan^2x+\sec^2x\]
AakashSudhakar
  • AakashSudhakar
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!
UsukiDoll
  • UsukiDoll
I think there's already an identity which is \[\tan^2x+1 =\sec^2x \] then just rearrange the terms.
UsukiDoll
  • UsukiDoll
oh I see where you're going on this... xD
AakashSudhakar
  • AakashSudhakar
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!
anonymous
  • anonymous
im confused now :/
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
\(\color{blue}{\text{Originally Posted by}}\) @UsukiDoll there is a trig identity which is \[\tan^2x+1 =\sec^2x \] \(\color{blue}{\text{End of Quote}}\)
AakashSudhakar
  • AakashSudhakar
@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.
UsukiDoll
  • UsukiDoll
anonymous
  • anonymous
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
UsukiDoll
  • UsukiDoll
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
AakashSudhakar
  • AakashSudhakar
Unfortunately, I'm not quite sure what that means. Does it want both sides of the equation to be equal?
anonymous
  • anonymous
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.
UsukiDoll
  • UsukiDoll
\[-\tan^2x+\sec^2x=1 \] using the identity \[\tan^2x+1 =\sec^2x \] subtract \[\tan^2x \] from both sides \[1 =\sec^2x-tan^2x \] rearrange \[1 =-tan^2x+sec^2x \]
anonymous
  • anonymous
is that the final answer?
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
okay, i see now.
UsukiDoll
  • UsukiDoll
ugh my latex broke
UsukiDoll
  • UsukiDoll
but we got them to equal.. that's the main thing
anonymous
  • anonymous
i understand now, thank you sooo much!!

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