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Math Help - Nid help please

  1. #1
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    Nid help please

    Prove the identities:
    1.) cot (A + B) = (cotAcotB - 1) / (cotB + cotA)
    2.) cot (A - B) = (cotAcotB + 1) / (cotB - cotA)
    3.) sec (A + B) = (secAsecB) / (1 - tanAtanB)
    4.) csc (A + B) = (cscAcscB) / (cotB + cotA)
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    Prove the identities:
    1.) cot (A + B) = (cotAcotB - 1) / (cotB + cotA)
    \frac{\cos (x+y)}{\sin (x+y)}
    \frac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\sin y\cos x}
    Divide numerator and denominator by \sin x\sin y, thus,
    \frac{\cot x\cot y-1}{\cot y+\cot x}
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    2.) cot (A - B)
    Simple,
    \cot (x-y)=\cot (x+(-y))
    \frac{\cot x\cot (-y)-1}{\cot x+\cot (-y)}
    \frac{-\cot x\cot y-1}{\cot x-\cot y}
    Multiply numerator and denominator by (-1):
    \frac{\cot x\cot y+1}{\cot y-\cot x}
    Last edited by ThePerfectHacker; August 27th 2006 at 06:13 AM.
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    3.) sec (A + B) = (secAsecB) / (1 - tanAtanB)
    \frac{1}{\cos (x+y)}=\frac{1}{\cos x\cos y-\sin x\sin y}
    Divide numerator and denominator by \cos x\cos y
    Thus,
    \frac{\frac{1}{\cos x\cos y}}{1-\tan x\tan y}

    \frac{\sec x\sec y}{1-\tan x\tan y}
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  5. #5
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    4.) csc (A + B) = (cscAcscB) / (cotB + cotA)
    \frac{1}{\sin (x+y)}=\frac{1}{\sin x\cos y+\cos x\sin y}
    Divide numerator and denominator by \sin x\sin y
    Thus,
    \frac{\frac{1}{\sin x\sin y}}{\cot y+\cot x}

    \frac{\csc x\csc y}{\cot x+\cot y}
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  6. #6
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    Quote Originally Posted by ThePerfectHacker
    Simple,
    \cot (x-y)=\cot (x-(-y))
    \frac{\cot x\cot (-y)-1}{\cot x+\cot (-y)}
    \frac{-\cot x\cot y-1}{\cot x-\cot y}
    Multiply numerator and denominator by (-1):
    \frac{\cot x\cot y+1}{\cot y-\cot x}
    got a q sir
    how come cot (x-y) became equal to cot (x-(-y)) ?
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  7. #7
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    got a q sir
    how come cot (x-y) became equal to cot (x-(-y)) ?
    It didn't. Its a typo. It should read cot(x - y) = cot(x + (-y))

    Then he uses the answer from part (a)
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  8. #8
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    Problem Fixed, thank you for noticing.
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