If $\frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}$, then $\sin(b-c)=0$ or $\sin 2a + \sin2b + \sin 2c=0$ If $$\frac{\tan (\alpha + \beta - \gamma )}{\tan (\alpha - \beta + \gamma )}=\frac{\tan \gamma}{\tan \beta}$$
then prove 
$$\sin (\beta - \gamma)=0$$
or $$\sin 2\alpha + \sin2\beta + \sin 2\gamma =0$$
here my try goes -
To prove $\sin (\beta - \gamma)=0$, sin x = 0, only when x = 0 or 360 degree, but for the sake of simplicity let's take x = 0, then we've to prove that $\beta - \gamma =0$ or $\beta = \gamma$.
Now, 
If $\beta = \alpha$, then first equation will be $\frac{\tan \alpha}{\tan \alpha}= \frac{\tan\gamma}{\tan\beta}$ 
or $\frac{\tan\gamma}{\tan\beta}=1$
and 
$$\frac{\tan x}{\tan y}=1$$, only if $$\tan x = \tan y$$
So, also in this case $\tan \gamma$ must be equal to $\tan \beta$ so, as to prove that $\sin (\beta - \gamma)=0$.
So, how to prove that $\beta = \gamma$?
and how to solve the second part of the question here?
 A: If I add $1$ to both sides of your first equation then exploit sum-to-product formulas, I'll get$$\frac{\sin2\alpha}{\sin(\alpha-\beta+\gamma)\cos(\alpha+\beta-\gamma)}=\frac{2\sin2\alpha}{\sin2\alpha\cos(-2\beta+2\gamma)}=\frac{\sin(\beta+\gamma)}{\sin\beta\cos\gamma}.$$ If I instead subtract $1$ and then re-evaluate, I'll have $$-\frac{\sin(2\gamma-2\beta)}{\cos(\alpha+\beta-\gamma)\sin(\alpha-\beta+\gamma)}=-\frac{\sin(\gamma-\beta)\cos(\gamma-\beta)}{\cos(2\gamma-2\beta)\sin2\alpha}=-\frac{\sin(\beta-\gamma)}{\cos\gamma\sin\beta}.$$
Dividing the two equalities leaves\begin{align}-\frac{\sin2\alpha}{\sin(2\gamma-2\beta)}=\frac{\sin(\beta+\gamma)}{\sin(\gamma-\beta)}\rightarrow-&\sin(\gamma-\beta)\sin2\alpha=2\sin(\beta+\gamma)\sin(\gamma-\beta)\cos(\gamma-\beta)\\
&=\sin(\gamma-\beta)\big(\sin2\gamma+\sin2\beta\,\big),
\end{align}so either $\sin(\gamma-\beta)=0$ or $0=\sin2\alpha+\sin2\beta+\sin2\gamma$.
A: Let's define $\mu := \frac12(\beta+\gamma)$ and $\nu:=\frac12(\beta-\gamma)$, so that we have
$$\frac{\tan(\alpha+2\nu)}{\tan(\alpha-2\nu)} = \frac{\tan(\mu-\nu)}{\tan(\mu+\nu)} \quad\to\quad
\frac{\sin(\alpha+2\nu)\cos(\alpha-2\nu)}{\sin(\alpha-2\nu)\cos(\alpha+2\nu)}=\frac{\sin(\mu-\nu)\cos(\mu+\nu)}{\sin(\mu+\nu)\cos(\mu-\nu)}$$
The identity
$$\sin(p\pm q)\cos(p \mp q) = \sin p \cos p \pm \sin q \cos q = \frac12( \sin 2p \pm \sin 2q) \tag{$\star$}$$
gives us
$$\frac{\sin 2\alpha + \sin 4\nu}{\sin 2\alpha - \sin 4 \nu} 
= \frac{\sin 2\mu - \sin 2 \nu }{\sin 2 \mu + \sin 2\nu}$$ 
so that, upon cross-multiplying and combining terms,
$$\begin{align}
0 &= \sin 2\alpha \cdot 2 \sin 2\nu + \sin 4\nu \cdot 2\sin 2\mu \\
&= \sin 2\alpha \cdot 2 \sin 2\nu + 2\sin 2\nu \cos 2\nu \cdot 2\sin 2\mu \\  
&= 2\sin 2\nu\left(\; \sin 2\alpha + 2\sin2\mu\cos 2\nu \;\right) \\  
&= 2\sin(\beta-\gamma)\left(\; \sin 2\alpha + 2\sin(\beta+\gamma)\cos(\beta-\gamma) \;\right) \\  
&= 2\sin(\beta-\gamma)\left(\; \sin 2\alpha + \sin 2 \beta + \sin 2\gamma  \;\right) \qquad\text{(using $(\star)$ again)} \\  
\end{align}$$
Therefore,
$$\sin(\beta-\gamma) = 0 \qquad\text{or}\qquad \sin 2\alpha + \sin 2 \beta + \sin 2\gamma = 0$$
$\square$
A: $$\dfrac{\tan C}{\tan B}=\dfrac{\tan(A+(B-C))}{\tan(A-(B-C))}$$
$$\iff\dfrac{\sin C\cos B}{\cos C\sin B}=\dfrac{\sin(A+(B-C))\cos(A-(B-C))}{\sin(A-(B-C))\cos(A+(B-C))}$$
Using Componendo et Dividendo,
$$\dfrac{\sin(B+C)}{\sin(B-C)}=\dfrac{\sin2A}{-\sin2(B-C)}$$
$$\implies\sin(B-C)\sin2A=-\sin(B+C)\sin2(B-C)$$
The right hand side can be simplified as
$$=-\sin(B-C)\{2\sin(B+C)\cos(B-C)\}$$
$$=-\sin(B-C)(\sin2B+\sin2C)$$
