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Math Help - verifying idenitity

  1. #1
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    verifying idenitity

    how would i verify this identity?
    (1+sec(-x))/(sin(-X)+tan(-x))= -csc x
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  2. #2
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    Hello wolfxpax
    Quote Originally Posted by wolfxpax View Post
    how would i verify this identity?
    (1+sec(-x))/(sin(-X)+tan(-x))= -csc x
    First you need to know how to handle the negative signs:

    • \sin(-x) = -\sin x
    • \cos(-x) = \cos x

    So it follows that \tan(-x) =-\tan x, \sec(-x) = \sec x, and so on.

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

    = \frac{1+\sec x}{-\sin x - \dfrac{\sin x}{\cos x}}

    = \frac{1+\sec x}{-\sin x - \dfrac{\sin x}{\cos x}}\times\color{red}\frac{\cos x}{\cos x}

    = \frac{\cos x + 1}{-\sin x\cos x - \sin x}

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

    = \frac{1}{-\sin x}

    =-\csc x

    Grandad
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  3. #3
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    Hello, wolfxpax!

    A slightly different approach . . .


    \frac{1+\sec(\text{-}x)}{\sin(\text{-}x)+\tan(\text{-}x)} \:=\: -\csc x
    As Grandad pointed out: . \begin{array}{ccc} \sec(\text{-}x) &=& \sec(x) \\ \sin(\text{-}x) &=& \text{-}\sin(x) \\ \tan(\text{-}x) &=& \text{-}\tan(x) \end{array}


    So we have: . \frac{1+\sec x}{\text{-}\sin x - \tan x} \;=\;\frac{1+\sec x}{\text{-}\sin x  - \frac{\sin x}{\cos x}} \;=\;\frac{1+\sec x}{\text{-}\sin x\left(1 + \frac{1}{\cos x}\right)}

    . . . . . . . = \;\frac{1 + \sec x}{\text{-}\sin x(1 + \sec x)} \;=\;\frac{1}{-\sin x} \;=\;-\csc x

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