# Trigonometry help

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

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

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?

mjsd
Homework Helper
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: $$\frac{a}{c}+\frac{c}{a} = \frac{a^2+c^2}{a c}$$

ok ok so then it equals c^2+a^2/ca
I still don't see how the top can equal b^2

mjsd
Homework Helper
did someone mentioned that this triangle is right-angled?

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

i thought c was always the hypotenuse ...so b^2=c^2-a^2

cristo
Staff Emeritus