anonymous
  • anonymous
sinx/1-cosx+sinx/1+cosx=cscx
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
sinx/1-cosx+sinx/1+cosx=cscx
anonymous
  • anonymous
is it \[{\sin x \over 1-\cos x}+{\sin x \over 1+\cos x}=\csc x \]?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
\[={\sin x(1+\cos x)+\sin x (1-\cos x) \over 1-\cos^2 x}\] \[={ \sin x (1+\cos x+1 -\cos x) \over \sin^2 x}\] \[={2 \over \sin x}=2\csc x\]
anonymous
  • anonymous
are you sure the right hand side of the equation is cscx? because it should be 2 cscx!! assuming the question is asking to prove the identity.
anonymous
  • anonymous
i also got 2cscx . are you asked to prove the equation? if so, the equation can't be proved because it's wrong
anonymous
  • anonymous
no your right i miss type it thank you!
anonymous
  • anonymous
no problem
anonymous
  • anonymous
tan^2x=sec^2x-sin^2x-cos^2x ok its the same thing
anonymous
  • anonymous
this is an easy one.. \[=\sec^2 x-(\sin^2 x+\cos^2 x)= \sec^2 x-1 = \tan^2 x\] so you have both sides of the equation equal
anonymous
  • anonymous
tan^2x = sec^2x - (sin^2x + cos^2x) tan^2x = sec^2x - 1 tan^2x = (1/cos^2x) -1 tan^2x = (1-cos^x)/cos^2x tan^2x = sin^2x/cos^2x
anonymous
  • anonymous
sinx/cosx = tanx so -> sin^2x/cos^2x will be equal to tan^2x
anonymous
  • anonymous
sec^4x-tan^4x=sec^2x+tan^2x last one in this group
anonymous
  • anonymous
\[\sec^4 x-\tan^4 x=(\sec^2 x+\tan^2 x)(\sec^2 x- \tan^2 x)\] since sec^2-tan^2x=1, \[=\sec^2 x+\tan^2x\]
anonymous
  • anonymous
a fire is sighted due west of lookout A. the bearing of the fire from lookout B, 12.9 miles due south of A, is N 30^o 36' W. How far is the fire from B

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