Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook