LynFran
  • LynFran
please help me prove this identity im stuck not geting it out at all! (1+tanx+cotx)(cosx-sinx)=(cscx/sec^2x -secx/csc^2x)
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Loser66
  • Loser66
just distribute the left hand side, you can get the right hand side.
LynFran
  • LynFran
im ending up with \[(-\sin^2x/cosx+\cos^2x/sinx)\] when i distribute it
Loser66
  • Loser66
yyyyyyyyyyyyyyyyyyyes

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Loser66
  • Loser66
or \(\dfrac{cos^2}{sin}-\dfrac{sin^2}{cos}\) right?
Loser66
  • Loser66
now, \(sin = \dfrac{1}{csc}\) hence \(csc = \dfrac{1}{sin}\) right?
Loser66
  • Loser66
and \(sec (x) =\dfrac{1}{cos(x)} \rightarrow cos(x) =\dfrac{1}{sec(x)}\)
Loser66
  • Loser66
hence the first term is \(\dfrac{csc (x)}{sec^2(x)}\) do the same with the second term. combine to get the right hand side
Loser66
  • Loser66
got it?
LynFran
  • LynFran
i think so thanks
Loser66
  • Loser66
ok

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