If $\displaystyle \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}$
Prove that $\displaystyle \sin2a+\sin2b+\sin2c=0$
Hi
Let $\displaystyle a=b=c=\frac{\pi}{4}$
Then $\displaystyle \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan\left(\frac{\pi}{4}\right)}{\tan\l eft(\frac{\pi}{4}\right)}=1$
And $\displaystyle \frac{\tan c}{\tan b} = 1$
Therefore $\displaystyle \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}$
But $\displaystyle \sin2a+\sin2b+\sin2c=1+1+1=3$