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Trigonometry help

  1. Feb 15, 2007 #1
    ABC is right angled at B.

    Prove that tanA+tanC=b^2 / ac

    Ok so far i've got that tan A= a/c and tanC=c/a and then when i try to add those together i get (c^2)(a^2)/ac , but how can (c^2)(a^2) = b^2
    all I've thought of is pythagorus's theorem. But for it to equal b^2, a^2 has to be taken away from c^2
  2. jcsd
  3. Feb 15, 2007 #2
    Are you sure about this? How did adding become multiplication?

    It doesn't

    Again are you sure? The Pythagorean theorem says that a2+b2=c2 where c is the hypotenuse of the right triangle and a and b are the two legs of the triangle. Is that the case in this problem?
  4. Feb 15, 2007 #3


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    Note: [tex]\frac{a}{c}+\frac{c}{a} = \frac{a^2+c^2}{a c}[/tex]
  5. Feb 15, 2007 #4
    ok ok so then it equals c^2+a^2/ca
    I still don't see how the top can equal b^2
  6. Feb 15, 2007 #5


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    did someone mentioned that this triangle is right-angled?
  7. Feb 15, 2007 #6
    by Pythagorean theorem, a2 + c2 = b2 for the triangle you mentioned. B is the right angle here. so b is the hypotenuse. so it's square should be equal to the sum of the squares of the other two sides, namely a and c
  8. Feb 15, 2007 #7
    i thought c was always the hypotenuse ...so b^2=c^2-a^2
  9. Feb 16, 2007 #8


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    It depends what you call the sides! In your first post, you appear to be adopting the naming that calls the sides opposite angles A, B and C, a, b and c respectively.

    Note that Pythagoras' Theorem simply says that the square of the hypotenuse is equal to the sum of the squares of the other two sides. Thus, in this case, b2=a2+c2
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