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Math Help - Solving a trig identity

  1. #1
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    Solving a trig identity

    LS=(tanx-tany)/(cotx-coty)

    LS=-tanxtany

    If possible try to avoid use of the tan compound formula.

    Thanks!
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  2. #2
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    What are you trying to do?
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  3. #3
    MHF Contributor red_dog's Avatar
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    \dfrac{\tan x-\tan y}{\cot x-\cot y}=\dfrac{\tan x-\tan y}{\dfrac{1}{\tan x}-\dfrac{1}{\tan y}}=

    =\dfrac{\tan x-\tan y}{\dfrac{\tan y-\tan x}{\tan x\tan y}}=\left(\tan x-\tan y\right)\cdot\dfrac{\tan x\tan y}{-\left(\tan x-\tan y\right)}=-\tan x\tan y
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    Hey where is the R.H.S. and what you trying to prove?
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  5. #5
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    Quote Originally Posted by youngb11 View Post
    LS=(tanx-tany)/(cotx-coty)

    LS=-tanxtany

    If possible try to avoid use of the tan compound formula.

    Thanks!
    \displaystyle\frac{tanx-tany}{cotx-coty}=\left(\frac{tanx}{tanx}\right)\frac{tanx-tany}{cotx-coty}=\frac{tanx(tanx-tany)}{1-tanxcoty}

    =\displaystyle\left(\frac{tany}{tany}\right)\frac{  tanx(tanx-tany)}{1-tanxcoty}=\frac{tanxtany(tanx-tany)}{tany-tanx}

    =\displaystyle\ -\frac{tanxtany(tany-tanx)}{tany-tanx}
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