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Inverse trigonometry problem

  1. Jun 2, 2016 #1
    1. The problem statement, all variables and given/known data

    After having struggled yesterday with this as much as I could, I am posting this problem here-
    if ##ax+bsec(tan^{-1}x)=c## and ##ay+bsec(tan^{-1}y)=c##, then prove that ##\frac{x+y}{1-xy}=\frac{2ac}{a^2-c^2}##.


    2. Relevant equations


    3. The attempt at a solution

    My attempt- Comparing both the equations, ##x=y## clearly(it is a dummy variable, sort of). So we basically need to find ##\frac{2x}{1-x^2}##. Put ##x=tan\theta##; so the given equation becomes-

    ##a tan\theta +b sec\theta=c##. From here, I don't know what to do. I tried to put it in the form of ##a sin\theta -(c) cos\theta=-b##, then divide and multiply it with ##\sqrt{(a^2+c^2)}## and put it in the auxiliary form(using ##a=rcos\theta,c=rsin\theta##); but alas; no help. By just working backwards, we see that the transformations ##a=rcos\theta## and ##c=rsin\theta## give the answer; and these are also the transforms I need to get my auxiliary form I mentioned above. But how do I connect them beats me. and Please help; I'd really appreciate it. Thanks in advance!
     
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  3. Jun 2, 2016 #2

    haruspex

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    Do you know a relationship between tan and sec?
     
  4. Jun 2, 2016 #3
    I doubt that it is necessarily true that x=y, that will hold only if one proves that the equation ##ax+bsec(tan^{-1}x)=c## has unique solution and I doubt a trigonometric equation has unique solution.

    From what i can see from wolfram solver, the equation does not has unique solution in real numbers, at least not in all cases.

    http://www.wolframalpha.com/input/?i=ax+bsec(arctan(x))=c
     
    Last edited: Jun 2, 2016
  5. Jun 3, 2016 #4
    @Delta² and @haruspex I found the answer-make the equation a quadratic with roots being x and y; then use properties of sum and products. OR, make an equation with an auxiliary angle, and then solve the resulting trig equation. Thanks for the time though!
     
  6. Jun 3, 2016 #5

    haruspex

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    Ok.
    Fwiw, I was hinting at using sec2=1+tan2, from which sec(tan-1(x))=√(1+x2). Maybe that's what you did.
     
  7. Jun 3, 2016 #6
    @haruspex yes that's what I did......and even going by my initial attempt as stated in the problem, I got the same result.
     
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