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Math Help - Prove:(tanx/(1+tanx)) = (sinx/(sinx+cosx))

  1. #1
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    Prove:(tanx/(1+tanx)) = (sinx/(sinx+cosx))

    For this question the only answer I seem to get is sinx/cosx not (sinx/(sinx+cosx))

    (tanx/(1+tanx)) = (sinx/(sinx+cosx))

    ((1-tanx/1-tanx)) times (tanx/(1+tanx))

    ((tanx(1-tanx))/(1-tan^2x))

    (tanx-tan^2x)/(1-tan^2x))

    ((cosx/cosx) times (sinx/cosx) - (sin^2x/cos^2x)) / ((cos^2x/cos^2x) times 1- (sin^2x/cos^2x))

    ((cosxsinx - sin^2x)/cos^2x) / ((cos^2x-sin^2x)/cos^2x)

    multiply the reciprocal and cancel i get

    sinx/cosx

    ugh.. what did i do wrong??
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  2. #2
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    you're really making this too hard ...

    \displaystyle \frac{\tan{x}}{1+\tan{x}} =

    \displaystyle\frac{\frac{\sin{x}}{\cos{x}}}{1 + \frac{\sin{x}}{\cos{x}}}

    finish it by multiplying numerator and denominator by \cos{x} ...
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  3. #3
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    ohh.. haha. thanks.. for some reason i never see the easiness of it .. ugh
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  4. #4
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    One of the easiest way to prove this we can also use it as (1-tanx)/tanx= (sinx-cosx)/cosx. because we changing both numerator so this will be ok and both sides will give you result cotx-1 so this is proved. Thanks
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  5. #5
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    Quote Originally Posted by tmas View Post
    For this question the only answer I seem to get is sinx/cosx not (sinx/(sinx+cosx))

    (tanx/(1+tanx)) = (sinx/(sinx+cosx))

    ((1-tanx/1-tanx)) times (tanx/(1+tanx))

    ((tanx(1-tanx))/(1-tan^2x))

    (tanx-tan^2x)/(1-tan^2x))

    ((cosx/cosx) times (sinx/cosx) - (sin^2x/cos^2x)) / ((cos^2x/cos^2x) times 1- (sin^2x/cos^2x))

    ((cosxsinx - sin^2x)/cos^2x) / ((cos^2x-sin^2x)/cos^2x)

    multiply the reciprocal and cancel i get

    sinx/cosx

    ugh.. what did i do wrong??
    You went wrong on your final step.

    \displaystyle\frac{\left(cosx\;sinx-sin^2x\right)\frac{1}{cos^2x}}{\left(cos^2x-sin^2x\right)\frac{1}{cos^2x}}=\frac{cosx\;sinx-sin^2x}{cos^2x-sin^2x}

    =\displaystyle\frac{sinx(cosx-sinx)}{(cosx-sinx)(cosx+sinx)}


    Simplest is to ask "how do we get from tanx\rightarrow\ sinx in the numerator" ?

    \displaystyle\frac{tanx}{1+tanx}=\left[\frac{cosx}{cosx}\right]\frac{tanx}{1+tanx}
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  6. #6
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    Whenever you are trying to prove a trig identity with tan x, cot x, sec x or csc x, a method that usually works pretty easily is to change these into sin x and cos x and then simplify algebraically. So if you don't see a quicker way right away always try this.
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  7. #7
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    Quote Originally Posted by ericpepin View Post
    One of the easiest way to prove this we can also use it as (1-tanx)/tanx= (sinx-cosx)/cosx. because we changing both numerator so this will be ok and both sides will give you result cotx-1 so this is proved. Thanks
    ok ... but what does that have to do with the original identity to be proved in this thread?
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  8. #8
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    Quote Originally Posted by skeeter View Post
    ok ... but what does that have to do with the original identity to be proved in this thread?
    I think here we trying to prove L.H.S.=R.H.S.
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