anonymous
  • anonymous
Verifying Trigonometric Identies help.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ 1 }{sinx-1 }-\frac{ 1 }{ sinx+1 }=-2\sec^2x\]
anonymous
  • anonymous
\[\frac{ sinx+1 }{\sin^2x-1}-\frac{ sinx-1 }{\sin^2x-1 }=\frac{ 2 }{ \sin^2x-1}\]
rishavraj
  • rishavraj
\[(a + b) (a - b) = (a^2 - b^2)\]

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anonymous
  • anonymous
that's what I got so far
rishavraj
  • rishavraj
u mean \[\frac{ \sin x + 1 - (\sin x -1) }{ \sin^2 x - 1 }\]
anonymous
  • anonymous
it wouldn't just be \[\frac{ 2 }{\sin^2x-1 }\]
rishavraj
  • rishavraj
so it would be \[\frac{ 2 }{ \sin^2 x - 1 }\] and cos^2 x + sin^2 x = 1 proceed
rishavraj
  • rishavraj
\[\sin^2 - 1 = - \cos^2 x \]
anonymous
  • anonymous
so \[\frac{ 2}{ \cos^2x }\]
rishavraj
  • rishavraj
its \[\frac{ -2 }{ \cos^2 x }\]
anonymous
  • anonymous
ah yeah I see that now
rishavraj
  • rishavraj
and \[\frac{ 1 }{ \cos x } = \sec x\]
anonymous
  • anonymous
I know sec x= 1/cos x
anonymous
  • anonymous
it still keeps the ^2 ?
rishavraj
  • rishavraj
\[\frac{ 1 }{ \cos^2 x } = \sec^2 x\]
anonymous
  • anonymous
thank you I got it now
rishavraj
  • rishavraj
u welcome :))

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