Solve equation on mathematical physics

Question

Show that $$\gamma_+ - \gamma_-=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}$$ where $$\gamma_+=(1-\beta^2_+)^{-\frac{1}{2}} \ \mbox{and} \ \beta_+=\frac{\beta_0+\beta}{1+\beta_0\beta}$$ $$\gamma_-=(1-\beta^2_-)^{-\frac{1}{2}} \ \mbox{and} \ \beta_-=\frac{\beta_0-\beta}{1-\beta_0\beta}$$

Answer 1

I was moved $\gamma_-$ to other side equation: $$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\gamma_-$$ So $$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\frac{1}{\sqrt{1-\beta^2_-}}$$ Therefore $$\gamma_-=\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}}$$ So we come to the conclusion that $$\gamma_-^2=\left(\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}}\right)^2$$ $$\gamma_-^2=\frac{\left(2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}\right)^2}{\left(\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}\right)^2}$$ Well I think?