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    #1

    Apr 14, 2005, 05:08 AM
    Trigonometric ratios
    Prove that (cot a+cosec a)^2=(1+cos a)/(1-cos a)
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    Dr_Calculus Posts: 35, Reputation: 1
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    #2

    Apr 15, 2005, 09:33 PM
    let's see...
    Well, [cot(x)+csc(x)]^2=cot(x)^2 + 2cot(x)csc(x) +csc(x)^2
    now, let's sub in sin and cos in place of cot and csc.

    cos^2/sin^2 +2(cos/sin)(1/sin)+1/sin^2

    now, we add it up because denominators are all sin^2.

    (cos^2+2cos+1)/sin^2 = (cos(x)+1)^2/(1-cos^2)

    note that I substituted the trig identity sin^2+cos^2=1 This in turn simplifies to:

    (cos(x)+1)^2/[(1-cos)(1+cos)]

    =(cos(x)+1)/(1-cos(x))

    It's a neat little trick.

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