# Trigonometry help

1. Feb 15, 2007

### wellY--3

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. Feb 15, 2007

### d_leet

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?

3. Feb 15, 2007

### mjsd

Note: $$\frac{a}{c}+\frac{c}{a} = \frac{a^2+c^2}{a c}$$

4. Feb 15, 2007

### wellY--3

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

5. Feb 15, 2007

### mjsd

did someone mentioned that this triangle is right-angled?

6. Feb 15, 2007

### murshid_islam

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. Feb 15, 2007

### wellY--3

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

8. Feb 16, 2007

### cristo

Staff Emeritus
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