anonymous
  • anonymous
Prove:
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[Tan^2x - \cos^2x = 1/\cos^2 - 1 - \cos^2x\]
lgbasallote
  • lgbasallote
can you rewrite that right side? is that \[\frac 1{cos^2 x - 1} - \cos ^2 x\]
anonymous
  • anonymous
|dw:1352207602501:dw|

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lgbasallote
  • lgbasallote
you know this identity right? \[\sin^2 x = 1 - \cos^2 x\]
anonymous
  • anonymous
yes
lgbasallote
  • lgbasallote
wait..nevermind what i said....
anonymous
  • anonymous
alright
lgbasallote
  • lgbasallote
\[\frac 1{\cos^2x} - 1 - \cos^2 x\] convert to common denominators first \[\implies \frac 1{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} - \frac{\cos^4 x}{\cos^2 x}\] agree?
anonymous
  • anonymous
yes :) I was on the right track whew :P
lgbasallote
  • lgbasallote
now combine the numerators \[\frac{1 - \cos^2 x - \cos ^ 4 x}{\cos^2 x}\] is this what you did?
anonymous
  • anonymous
yes but Instead if making it cos^4x I cancelled out the top bracket Cos^2x with the bottom one...which is probably where I went wrong?
lgbasallote
  • lgbasallote
no. that works too \[\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x\] yes?
anonymous
  • anonymous
hmm, why didnt you just totally cancel out the denominator?
anonymous
  • anonymous
I was left with 1-cos^2x-cos^2x :S
lgbasallote
  • lgbasallote
ah. that's different
lgbasallote
  • lgbasallote
you can't do that
anonymous
  • anonymous
|dw:1352208030577:dw|
anonymous
  • anonymous
hmm whats the reason I couldnt?
lgbasallote
  • lgbasallote
observe what i did earlier \[\frac 1{\cos^2 x} - 1 - \cos^2 x \implies \frac 1{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} - \frac{\cos^4 x}{\cos^2 x}\] each term has its own denominator
anonymous
  • anonymous
|dw:1352208100756:dw|
anonymous
  • anonymous
ap calc etcher tell me that
anonymous
  • anonymous
OH I understand
lgbasallote
  • lgbasallote
if you cancel out the cos^2 x below cos^4 x, you can only do it there, and not in the other terms. make sense?
anonymous
  • anonymous
yes! I love you :P
lgbasallote
  • lgbasallote
so anyway, the RHS simplifies to \[\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x\] i assume you know how to proceed form tehre?
anonymous
  • anonymous
should i move to the left side?
anonymous
  • anonymous
wo think i got it? L.S) will equel 1-cos^2/cos^2x - Cos^2x
anonymous
  • anonymous
Ls = Rs Sweet! Thanks for your help! I also found a few more I am having trouble with...this next question has a tan^3x which I haven't used in a while i forget what i can do with it
lgbasallote
  • lgbasallote
tan^3 x = tan^2 x * tan x
lgbasallote
  • lgbasallote
that can work
lgbasallote
  • lgbasallote
or maybe tan^3 x = sin^3 x / cos^3 x
anonymous
  • anonymous
\[\tan ^{2}x - \cos^{2}x =\frac{1}{ \cos^2(x)} -1- \cos^{2}x \] \[\tan ^{2}x - \cos^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)} - \cos^{2}x \] which is \\[\tan ^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)} \] which is \[\tan ^{2}x =\frac{(\sin^{2}x) }{ \cos^2(x)}\] which satisfis the relation
anonymous
  • anonymous
Thanks guys, lemme try this other one and i cant get it ill post a new question
anonymous
  • anonymous
all the best :)

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