How to eliminate $\theta$? While doing a sum I was stuck in a particular step:
$$r_1= \frac{4a \cos \theta }{\sin^2 \theta}$$ and $$r_2=\frac{4a \sin \theta }{\cos^2 \theta}$$
How to eliminate $\theta$ ?

$$r_1\cdot r_2 = \frac{(4a)^2}{\sin \theta \cos \theta}$$
$$\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}=\frac{\sin \theta}{\cos \theta}$$
$$\frac{\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}}{r_1\cdot r_2 }=\frac{\sin^2 \theta}{(4a)^2}=r_1^{-\frac{4}{3}}r_1^{-\frac{2}{3}}$$
$$\frac{1}{\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}r_1\cdot r_2 }=\frac{\cos^2 \theta}{(4a)^2}=r_1^{-\frac{2}{3}}r_1^{-\frac{4}{3}}$$
$$\therefore r_1^{-\frac{4}{3}}r_2^{-\frac{2}{3}}+r_1^{-\frac{2}{3}}r_2^{-\frac{4}{3}}=\frac{1}{(4a)^2}$$ 
