anonymous
  • anonymous
prove using identities that ((sec^2)x)/(((sec^2)x)-1)=(csc^2)x
Mathematics
chestercat
  • chestercat
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bahrom7893
  • bahrom7893
Okay, so let me try this: did you mean: \[Sec^2(x)/(Sec^2x - 1) = Csc^2(x)\]
anonymous
  • anonymous
yea. sorry i don't know how get it like that on my computer
bahrom7893
  • bahrom7893
okay so now workin on it

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anonymous
  • anonymous
thank you so much
bahrom7893
  • bahrom7893
There are a few Sec and Csc identities, but I never remember them so im just gonna convert em to Sin and Cos: Sec(x) = 1/Cos(x) Csc(x) = 1/Sin(x)
bahrom7893
  • bahrom7893
\[[1/Cos^2x] / [ (1/Cos^2x)-1 ] = 1/Sin^2x\]
bahrom7893
  • bahrom7893
Now let's simplify this part: \[(1/Cos^2x) - 1\] = \[(1-Cos^2x)/Cos^2x\]
anonymous
  • anonymous
wait. how did you do that?
bahrom7893
  • bahrom7893
find the common denominator
bahrom7893
  • bahrom7893
1 = Cos^2 / Cos^2... get it?
anonymous
  • anonymous
oh . yea, sorry. i got it
bahrom7893
  • bahrom7893
k
bahrom7893
  • bahrom7893
So now you have to divide 1/Cos^2x by (1-Cos^2x)/Cos^2x which is the same as multiplying by its inverse
bahrom7893
  • bahrom7893
\[[1/Cos^2x] * [Cos^2x/(1-Cos^2x)] = 1/Sin^2x\]
bahrom7893
  • bahrom7893
Cancel out cos^2s
bahrom7893
  • bahrom7893
\[1/(1-Cos^2x) = 1/Sin^2x\]
bahrom7893
  • bahrom7893
Cross multiply:
anonymous
  • anonymous
wait. can't you just say 1 / (sin^2x) =csc^2x
anonymous
  • anonymous
and then be done? or is that skipping a step?
bahrom7893
  • bahrom7893
yeah, but i prefer keeping everything in sin and cosine
bahrom7893
  • bahrom7893
bcuz its easy to simplify
bahrom7893
  • bahrom7893
\[Sin^2x = 1 - Cos^2x\]
bahrom7893
  • bahrom7893
See that's proved. Sin^2x + Cos^2x = 1!!!!
bahrom7893
  • bahrom7893
get it?
anonymous
  • anonymous
yes. thank you so much!
bahrom7893
  • bahrom7893
k off to helpin someone else.. any questions: email me bahrom.cfo@gmail.com and if u liked my explanation, bcum a fan YAY!
anonymous
  • anonymous
deff will. that was so helpful. thanks again!

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