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Math Help - Prove (1+cos x/1+cot x) = tan x

  1. #1
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    Prove (1+cos x/1+cot x) = tan x

    So when I did this the proof I got was -tanx and i can't really see where i went wrong:

    1+tanx/1+cotx =tanx

    (1-cotx/1-cotx) (1+tanx/1+cotx)

    1+tanx-cotx-cotxtanx/ 1-cot^2x

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

    ((sin^2xcosx- sinxcos^2x/sinxcosx)/( (sin^2x/sin^2x) - (cos^2x/sin^2x)))

    ((sinxcosx(sinx-cosx)/sinxcosx)) / sin^2x-cos^2x)

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

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

    sin^2x/(sinx-cosx)

    - sinx/cosx

    -tanx

    Hopefully this makes sense ....
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  2. #2
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    btw , fix your post title ...

    \displaystyle \frac{1+\tan{x}}{1 + \cot{x}} \cdot \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}} =

    \displaystyle \frac{\sin{x}\cos{x} + \sin^2{x}}{\sin{x}\cos{x} + \cos^2{x}} =

    \displaystyle \frac{\sin{x}(\cos{x} + \sin{x})}{\cos{x}(\sin{x} + \cos{x})} =

    \displaystyle \frac{\sin{x}}{\cos{x}}
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  3. #3
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    \frac{1 + tanx}{1 + cotx} = \frac{tanx(1 + tanx)}{tanx(1 + cotx)}

    Simplify the denominator and proceed.
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