# Prove tthat tan 50 * tan 60 * tan 70 = tan 80

1. Feb 11, 2005

### no name

hiiiiiiiiiii
i am here with a very tough question
prove tthat
tan 50 * tan 60 * tan 70 = tan 80
cheers
abc

2. Feb 11, 2005

### Davorak

Are you sure I get
50*Tan[60]*Tan[70] = 19.538
Tan[80]=9.00365
If in radians and the below if in degrees:
50*Tan[60]*Tan[70] = 237.939
Tan[80]=5.67128

A play on words maybe I am not catching it anyone else?

Edit: oops, Thanks gerben.

Last edited: Feb 11, 2005
3. Feb 11, 2005

### gerben

tan 50 * tan 60 * tan 70 = tan 80 (in degrees)
Not
50 * tan 60 * tan 70 = tan 80

4. Feb 11, 2005

### Icebreaker

Last edited by a moderator: Feb 11, 2005
5. Feb 11, 2005

### AKG

Icebreaker, you simply show that the quantities are equal to within whatever error Google's calculator computes things. I think the proof should involve the use of the following identity (with x = 10) :

$$\tan (nx) = \frac{\tan [(n - 1)x] + \tan (x)}{1 - \tan [(n - 1)x]\tan (x)}$$

6. Feb 12, 2005

### chronon

We know tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))

So tan(60+x)=(tan(60)+tan(x))/(1-tan(60)tan(x)) -- all numbers in degrees

with a corresponding formula for tan(60-x). Multiplying the two, using tan(60)^2=3 and putting t=tan(x)

tan(60-x)tan(60+x)=(3+t^2)/(1-3t^2)

Now putting x=10 degrees, and so t=tan(10)
we have tan(30)=(3t-t^3)/(1-3t^2) and tan(60)=1/tan(30) so

tan(50)tan(60)tan(70)=[(3+t^2)/(1-3^t^2)]/[(3t-t^3)/(1-3t^2)]=1/t
=1/tan(10)=tan(80)