unicwaan
  • unicwaan
Can someone help me verify the identity? sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x
Mathematics
jamiebookeater
  • jamiebookeater
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BTaylor
  • BTaylor
so, this? \[\frac{\sin x}{1-\cos{x}} + \frac{\sin x}{1-\cos{x}} = 2\csc x\]
unicwaan
  • unicwaan
yes except the it's 1- cos and 1 + cos sorry
welshfella
  • welshfella
first convert the 2 fractions on left side to one fraction

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BTaylor
  • BTaylor
You want to use the identity \(1-\cos^2 x = \sin^2 x\). Multiply the 1 - cos x by (1 + cos x) on top and bottom, and you will have sin^2 x on the bottom.
unicwaan
  • unicwaan
Okay I understand that step BTaylpr but I don't know how the make the other fraction the same like what welshfella wants me to do
misty1212
  • misty1212
HI!!
BTaylor
  • BTaylor
\[\frac{\sin x}{1-\cos{x}} \cdot \frac{1+\cos x}{1+\cos x} + \frac{\sin x}{1+\cos{x}} \cdot \frac{1-\cos x}{1-\cos x} = 2\csc x\]
misty1212
  • misty1212
you can also just add, you will get it that way too
unicwaan
  • unicwaan
Oh the conjugate! I forgot about that!
misty1212
  • misty1212
\[\frac{b}{1+a}+\frac{b}{1-a}=\frac{b(1-a)+b(1+a)}{(1-a)(1+a)}\]
misty1212
  • misty1212
you get \[\frac{2b}{1-a^2}\]
unicwaan
  • unicwaan
Thank you I justed needed that step to get me started, I can figure the rest out :)
unicwaan
  • unicwaan
just*
BTaylor
  • BTaylor
no problem

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