Trigonometry help

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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
 

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  • #2
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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
Are you sure about this? How did adding become multiplication?

but how can (c^2)(a^2) = b^2
It doesn't

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
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?
 
  • #3
mjsd
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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
Note: [tex]\frac{a}{c}+\frac{c}{a} = \frac{a^2+c^2}{a c}[/tex]
 
  • #4
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ok ok so then it equals c^2+a^2/ca
I still don't see how the top can equal b^2
 
  • #5
mjsd
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did someone mentioned that this triangle is right-angled?
 
  • #6
ok ok so then it equals c^2+a^2/ca
I still don't see how the top can equal b^2
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
 
  • #7
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i thought c was always the hypotenuse ...so b^2=c^2-a^2
 
  • #8
cristo
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i thought c was always the hypotenuse ...so b^2=c^2-a^2
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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