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anonymous

  • 5 years ago

sinx/1-cosx+sinx/1+cosx=cscx

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  1. anonymous
    • 5 years ago
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    sinx/1-cosx+sinx/1+cosx=cscx

  2. anonymous
    • 5 years ago
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    is it \[{\sin x \over 1-\cos x}+{\sin x \over 1+\cos x}=\csc x \]?

  3. anonymous
    • 5 years ago
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    yes

  4. anonymous
    • 5 years ago
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    \[={\sin x(1+\cos x)+\sin x (1-\cos x) \over 1-\cos^2 x}\] \[={ \sin x (1+\cos x+1 -\cos x) \over \sin^2 x}\] \[={2 \over \sin x}=2\csc x\]

  5. anonymous
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    i also got 2cscx . are you asked to prove the equation? if so, the equation can't be proved because it's wrong

  7. anonymous
    • 5 years ago
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    no your right i miss type it thank you!

  8. anonymous
    • 5 years ago
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    no problem

  9. anonymous
    • 5 years ago
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    tan^2x=sec^2x-sin^2x-cos^2x ok its the same thing

  10. anonymous
    • 5 years ago
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    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

  11. anonymous
    • 5 years ago
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    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

  12. anonymous
    • 5 years ago
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    sinx/cosx = tanx so -> sin^2x/cos^2x will be equal to tan^2x

  13. anonymous
    • 5 years ago
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    sec^4x-tan^4x=sec^2x+tan^2x last one in this group

  14. anonymous
    • 5 years ago
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    \[\sec^4 x-\tan^4 x=(\sec^2 x+\tan^2 x)(\sec^2 x- \tan^2 x)\] since sec^2-tan^2x=1, \[=\sec^2 x+\tan^2x\]

  15. anonymous
    • 5 years ago
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    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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