# Proving two sides of equation for triangles

1. Nov 6, 2007

### rum2563

[SOLVED] Proving two sides of equation for triangles

1. The problem statement, all variables and given/known data
In angle ABC, which is an isosceles triangle with <B = <C, show that
2cot(a) = tan(b) = cot(b)

2. Relevant equations
tan2a = 2tana / 1 - tan^2 a

tan (x - y) = tanx - tany / 1 + tanx tany

3. The attempt at a solution

Since it is isosceles, that means two sides are equal, therefore, <A = 180 - 2B

2cot(A) = 2cot (180 - 2B)
= 1 / 2tan(180 - 2B)
= 1 / 2(tan180 - tan2B / 1 + tan180 tan2B)
= 1 / 2(-tan2B)
= 1 / -2tan2B)
= 1 / -2(2tanB / 1 - tan^2 B)

2. Nov 6, 2007

### learningphysics

2*cot(180 - 2B) = 2*[1/tan(180-2B)] = 2/tan(180-2B)

also, please uses parentheses... in your work here especially when dealing with fractions and sums in the numerator and denominator. people will get confused as to your exact meaning: for example this:

means: tan180 - tan2B + (tan180)*(tan2B) the way you wrote it.

you should write:

(tan180 - tan2B) / (1 + tan180 tan2B )

3. Nov 6, 2007

### rum2563

Thanks for your help. Here is what I am getting:

2cot(a) = 2/tan(180-2B)
= 2 / {(tan180 - tan2b) / (1 + tan180tan2b)}
= 2 / {-tan2b / 1}
= 2 / {-2tanb / 1 - tan^2 b)

After this, i am still confused as to how to get the equation to equal to tanb - cotb.

4. Nov 6, 2007

### learningphysics

2 / [-2tanb / (1 - tan^2 b)]

= $$2*(\frac{1-tan^2(b)}{-2tan(b)})$$

=$$\frac{1-tan^2(b)}{-tan(b)}$$

= -1/tan(b) + tan(b)
= -cot(b) + tan(b)
= tan(b) - cot(b)

5. Nov 6, 2007

### rum2563

Wow, thanks very much learningphysics. You have helped me so much and even shown me the answer. I was worried about the negative sign and how to get rid of it, but as soon as I saw your solution, I understood quickly the idea behind it. Thanks very much again.

6. Nov 6, 2007

### learningphysics

no prob. you're welcome.