# Working using trig identity

Hi so I have this equation

$$\frac{1-e^2}{1+e\cos(\theta -\theta_{0})} = 1-e\cos\eta$$

and using the identity

$$\cos\theta = \frac{1-\tan^2 \frac\theta2}{1+\tan^2 \frac\theta2}$$

this becomes

$$\sqrt{1-e} \tan\left(\frac{\theta - \theta_{0}}2\right) = \sqrt{1+e} \tan\frac\eta2.$$

However, I'm unsure about the steps in between. Can someone help me out please?

• @BarryCipra apologies, supposed to be a + on the bottom. edited it now! – SFL Apr 23 '18 at 20:45

Set $u=\tan((\theta-\theta_0)/2)$ and $v=\tan(\eta/2)$ for simplicity.
The left hand side becomes $$\frac{1-e^2}{1+e\dfrac{1-u^2}{1+u^2}} =\frac{(1-e^2)(1+u^2)}{(1+e)+(1-e)u^2}$$ The right hand side becomes $$1-e\frac{1-v^2}{1+v^2}=\frac{(1-e)+(1+e)v^2}{1+v^2}$$ It's better to set $M=1-e$ and $P=1+e$, so we get a simpler identity $$\frac{MP(1+u^2)}{P+Mu^2}=\frac{M+Pv^2}{1+v^2}$$ that can be rearranged to $$MP+MPv^2+MP(1+v^2)u^2=MP+P^2v^2+M^2u^2+MPu^2v^2$$ Pull all terms with $u^2$ on the left hand side $$(MP+MPv^2-M^2-MPv^2)u^2=P^2v^2-MPv^2$$ that simplifies to $$M(P-M)u^2=P(P-M)v^2$$ Assuming $P-M=1+e-1+e=2e\ne0$, this becomes $$(1-e)\tan^2\frac{\theta-\theta_0}{2}=(1+e)\tan^2\frac{\eta}{2}$$