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Math Help - Verifying Identities II

  1. #1
    nee
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    Verifying Identities II

    tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx

    HELPO!!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by nee View Post
    tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx

    HELPO!!!
    Move from LHS to RHS:

    \begin{aligned}\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}&=\frac{\tan x(1-\cos x)}{1-\cos^2x}+\frac{\sin x(1+\cos x)}{1-\cos^2x}\\&=\frac{(\tan x-\sin x)+(\sin x +\sin x\cos x)}{1-\cos^2x}\end{aligned}

    Can you finish this? If you can't, see the spoiler for the next hint.

    Spoiler:
    Apply the identity 1-\cos^2x=\sin^2x and simplify the result...
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  3. #3
    nee
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    (tanx - sinx) + (sinx + sinxcosx)
    -------------------------------
    sin^2x

    = tanx + sinxcosx ?


    How do you get the reciprocal identities at the end though. Shouldn't it simplify to cotx + secxcscx ?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by nee View Post
    (tanx - sinx) + (sinx + sinxcosx)
    -------------------------------
    sin^2x

    = tanx + sinxcosx ?


    How do you get the reciprocal identities at the end though. Shouldn't it simplify to cotx + secxcscx ?
    It should simplify to \cot x+\sec x\csc x...

    \begin{aligned}\frac{\tan x-\sin x+\sin x+\sin x\cos x}{\sin^2x}&=\frac{\tan x}{\sin^2 x}+\frac{\sin x\cos x}{\sin^2 x}\\ &=\frac{1}{\cos x\sin x}+\frac{\cos x}{\sin x}\\ &=\ldots\end{aligned}
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  5. #5
    nee
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    Oh ok, I simplified wrong. Thanks!!!
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