# Math Help - Prove:

1. ## Prove:

If $\frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}$

Prove that $\sin2a+\sin2b+\sin2c=0$

2. Originally Posted by great_math
If $\frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}$

Prove that $\sin2a+\sin2b+\sin2c=0$
Hi

Let $a=b=c=\frac{\pi}{4}$

Then $\frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan\left(\frac{\pi}{4}\right)}{\tan\l eft(\frac{\pi}{4}\right)}=1$

And $\frac{\tan c}{\tan b} = 1$

Therefore $\frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}$

But $\sin2a+\sin2b+\sin2c=1+1+1=3$

3. Hey that's not what you were supposed to prove!

4. Originally Posted by Bruno J.
Hey that's not what you were supposed to prove!
When something seems very difficult to prove I prefer spending some time trying to find a counter example showing that it is false !