Eliminating $r$ from $6\tan(r+x)=3\tan(r+y)=2\tan(r+z)$

Solution 1:

Just as you derive,

$$\sin(2r+x+y)=3\sin(y-x)\tag1$$ $$\sin(2r+y+z)=5\sin(z-y)\tag2$$

you can get, $$\sin(2r+z+x)=-2\sin(z-x)\tag3$$

From (1), can you get, $$\begin{align}\sin(2r+x+y)\sin(y-x)&=3\sin^2(y-x)\\ \frac12\left[\cos(2r+2x)-\cos(2r+2y)\right]&=3\sin^2(y-x)\tag{1´}\end{align}$$?

From (2), can you get, $$\begin{align} \sin(2r+y+z)\sin(z-y)&=5\sin^2(z-y)\\ \frac12\left[\cos(2r+2y)-\cos(2r+2z)\right]&=5\sin^2(z-y)\tag{2´}\end{align}$$?

Similarly, from (3), can you get, $$\begin{align} \sin(2r+z+x)\sin(x-z)&=-2\sin^2(z-x)\\ \frac12\left[\cos(2r+2z)-\cos(2r+2x)\right]&=-2\sin^2(z-x)\tag{3´}\end{align}$$?

Observe what happens when you add, $(1´)$, $(2´)$ and $(3´)$.

Additionally, you can solve this more general problem:

If $$\small \boldsymbol{a\tan (r+x)=b\tan(r+y)=c\tan(r+z)}$$ show that $$\small \boldsymbol{\left(\frac{a+b}{a-b}\right)\sin^2(x-y)+\left(\frac{b+c}{b-c}\right)\sin^2(y-z)+\left(\frac{c+a}{c-a}\right)\sin^2(z-x)=0}.$$

and enjoy! : )