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Math Help - Idenities

  1. #1
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    Question Idenities

    Hey does anyone know these trig idenities??



    ((Cosx +1)/(secx-tanx)) - ((1-cosx)/secx+tanx)) = 2(1+tanx)



    sec(-x) / sin(180-x) = tan(270-x) + tan(180+x)


    (tanx-sinx) / (sin(cubed)x) = (sec) / (1+cosx)
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  2. #2
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    Hello Giggly2
    Quote Originally Posted by Giggly2 View Post
    Hey does anyone know these trig idenities??
    If you mean do I know how to prove them, yes.
    ((Cosx +1)/(secx-tanx)) - ((1-cosx)/secx+tanx)) = 2(1+tanx)
    For this one, you need to know that if you divide both sides of \cos^2x +\sin^2x = 1 by \cos^2x, you get:

    1+\tan^2x = \sec^2x

    \Rightarrow \sec^2x-\tan^2x = 1

    \Rightarrow (\sec x+\tan x)(\sec x - \tan x) = 1

    Then just put both the fractions on the LHS over a common denominator, multiply out the brackets (be careful with the minus signs!) and you'll get 2(1 + \tan x). I'll start you off:

    \frac{\cos x +1}{\sec x -\tan x}-\frac{\cos x -1}{\sec x +\tan x} = \frac{(\cos x +1)(\sec x +\tan x)-(1-\cos x)(\sec x - \tan x)}{(\sec x+\tan x)(\sec x - \tan x)}

    = \frac{\dots \quad \dots \quad \dots \quad \dots}{1}

    Can you complete it now?


    sec(-x) / sin(180-x) = tan(270-x) + tan(180+x)
    You need to know the following identities:

    • \sec(-x) = \sec x
    • \sin (180 - x) = \sin x
    • \tan(180 + x) = \tan x
    • So \tan(270 - x) = \tan(180 + 90 - x)=\tan(90-x) = \cot x

    Then \frac{\sec(-x)}{\sin(180-x)}= \frac{\sec{x}}{\sin x} = \frac{1}{\cos x\sin x}

    Now

    • start from the RHS using the last two identities above
    • express \cot x and \tan x in terms of sine and cosine
    • add the two fractions
    • use \cos^2x +\sin^2x = 1

    and you're there.

    (tanx-sinx) / (sin(cubed)x) = (sec) / (1+cosx)
    Again, I'll start you off:

    \frac{\tan x - \sin x}{\sin^3x}= \frac{\frac{\sin x}{\cos x}-\sin x}{\sin^3x}

    = \frac{\sin x - \sin x \cos x}{\cos x\sin^3x}

    Now you need to:

    • factorise the numerator
    • divide top and bottom by the common factor \sin x
    • use \sin^2x = 1-\cos^2x =(1+\cos x)(1-\cos x)
    • divided top and bottom by a common factor again

    You should end up with \frac{1}{\cos x(1+\cos x)}= \frac{\sec x}{1+\cos x}

    Can you finish them off now?

    Grandad
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  3. #3
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    Thank you soooo much, that was alot of help!!
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