anonymous
  • anonymous
(sinx / one - cosx )+ (sin x /one - cosx) = 2 csc x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • Loser66
How??? check the problem, please.
Loser66
  • Loser66
but it is true if the denominator is \(1-cos^2(x)\)
anonymous
  • anonymous
what identity would i use @Concentrationalizing

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Loser66
  • Loser66
@Concentrationalizing please, give me the whole thing. Can't wait to see "the 2 more steps" to get the answer.
anonymous
  • anonymous
Ah, I see what you mean. I wasn't paying attention, I apologize.
anonymous
  • anonymous
so i would use the Pythagorean identity ? @Concentrationalizing
anonymous
  • anonymous
No, I was wrong. Loser is right, the problem is faulty. I didnt even bother to check ahead, I just went through the motions of what steps you might normally do.
anonymous
  • anonymous
so i'm confused what are the steps i have to do? @Concentrationalizing @Loser66
anonymous
  • anonymous
The problem is incorrect. You maybe typed something wrong, but your question isn't true.
anonymous
  • anonymous
As loser said, this is the set-up that would be correct http://www.wolframalpha.com/input/?i=sinx%2F%281-cos^2%28x%29%29+%2B+sinx%2F%281-cos^2%28x%29%29+%3D+2cscx+
anonymous
  • anonymous
@Concentrationalizing
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anonymous
  • anonymous
Okay, you typed it wrong in your original question then.
anonymous
  • anonymous
\[\frac{ sinx }{ 1-cosx }+\frac{ sinx }{ 1+cosx } = \frac{ sinx(1+cosx)+sinx(1-cosx) }{(1+cosx)(1-cosx) }\] \[\frac{ sinx + sinxcosx + sinx - sinxcosx }{ 1-\cos^{2}x }\] \[\frac{ 2sinx }{ 1-\cos^{2}x }\] And yes, pythagorean identity from here. \(sin^{2}x + cos^{2}x = 1\) can be rearranged into \(sin^{2}x = 1-cos^{2}x\) Thus we have \[\frac{ 2sinx }{ \sin^{2}x }= \frac{ 2 }{ sinx } = 2cscx\]
anonymous
  • anonymous
thank you @Concentrationalizing

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