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Math Help - Tangents

  1. #1
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    Tangents

    if x+y+z+t=\pi then \tan x+\tan y+\tan z+\tan t=\tan x\tan y\tan z\tan t
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  2. #2
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    Re: Tangents

    Quote Originally Posted by TheodorMunteanu View Post
    if x+y+z+t=\pi then \tan x+\tan y+\tan z+\tan t=\tan x\tan y\tan z\tan t
    Hint:

    \tan x+\tan y+\tan z+\tan t=

    =\frac{1}{4}\sec t\sec x\sec y\sec z \cdot(-\sin(t-x-y-z)+\sin(t+x+y-z)+\sin(z+x-y+z)+\sin(t-x+y+z)+2\sin(x+y+z+t))
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  3. #3
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    Re: Tangents

    Actually you're trying to prove a wrong exercise:
    Correct it would have been: \tan x+\tan y+\tan z+\tan t=\frac{\tan x+\tan y+\tan z+\tan t-\tan x\tan y\tan z-tan x\tan z \tan t-\tan x\tan y\tan t-\tan y\tan z\tan t}{\tan x\tan y+...-\tan x\tan y\tan z\tan t}
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  4. #4
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    Re: Tangents

    Hello, TheodorMunteanu!

    \text{if }x+y+z+t\,=\,\pi,

    \text{ then: }\:\tan x+\tan y+\tan z+\tan t\:=\:\tan x\tan y\tan z\tan t
    This is not true!
    \text{Let }x = y = z = t = \tfrac{\pi}{4}

    \text{We have: }\:x + y + z + t \:=\:\pi

    \text{Take tangents: }\:\tan(x+y + z + t) \;=\;\tan(0)

    . . . . . . . . \tan\bigg[(x+y) + (z+t)\bigg] \;=\;0

    . . . . . . . \frac{\tan(x+y) + \tan(z+t)}{1 - \tan(x+y)\tan(z+t)} \;=\;0


    Assume that the denominator is not zero.
    A fraction equals zero if its numerator equals zero.

    \text{We have: }\:\tan(x+y) + \tan(z+t) \;=\;0

    . . \frac{\tan x + \tan y}{1 - \tan x\tan y} + \frac{\tan z + \tan t}{1 - \tan z\tan t}\;=\;0

    \text{Then: }\,\frac{(\tan\!x +\tan\!y)(1-\tan\!z\tan\!t) + (\tan\!z + \tan\!t)(1 - \tan\!x\tan\!y)}{(1-\tan x\tan y)(1 - \tan z\tan t)} \;=\;0


    Assume that the denominator is not zero.
    A fraction equals zero if its numerator equals zero.

    (\tan\!x +\tan\!y)(1-\tan\!z\tan\!t) + (\tan\!z + \tan\!t)(1 - \tan\!x\tan\!y) \;=\;0

    \tan x - \tan x\tan z\tan t + \tan y - \tan y \tan z\tan t
    . . . + \tan z - \tan x \tan y\tan z + \tan t - \tan x\tan y \tan t \;=\;0


    Therefore:

    \tan x + \tan y + \tan z + \tan t

    . . =\;\tan x\tan y\tan z + \tan x\tan y\tan t + \tan x\tan z\tan t + \tan y\tan z\tan t

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  5. #5
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    Re: Tangents

    Yeah well as you can see I've corrected my previous post 2 hours earlier.
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