Question :

Eliminate $\theta$ and $\phi$ from the relation

$$\begin{align} p \cot^2\theta + q \cot^2\phi &= 1 &(1)\\ p \cos^2\theta + q \cos^2\phi &= 1 &(2)\\ p \sin\theta &= q\sin\phi &(3) \end{align}$$

Also find relation between $p$ and $q$.

I have tried different ways but unable to eliminate $\theta$ and $\phi$.

One method: If I subtract equations $(1)$ and $(2)$ then I got:

$$p\frac{\cos^4\theta}{\sin^2\theta} +q \frac{\cos^4\phi}{\sin^2\phi}=0$$

Please guide how to solve this.. thanks..

  • 1
    $\begingroup$ I edited to improve the formatting, and to remove the redundant equations. Please check my changes for errors. $\endgroup$ – Blue Jul 8 '13 at 11:21
  • $\begingroup$ @Blue, nice alignment! specially getting the equation tags, so neat. $\endgroup$ – Arjang Jul 8 '13 at 11:21
  • $\begingroup$ @Arjang: The tag alignment comes with just an extra & in the align environment. I agree: it's a neat trick. :) $\endgroup$ – Blue Jul 8 '13 at 11:23

HINT: $$p\sin\theta=q\sin\phi\implies p^2\sin^2\theta=q^2\sin^2\phi$$

$$\implies p^2\cos^2\theta-q^2\cos^2\phi=p^2-q^2\ \ \ \ (4)$$

Use $(2),(4)$ to solve for $\cos^2\phi,\cos^2\theta$

If $\cos^2\phi=y, \cot^2\phi=\frac{\cos^2\phi}{1-\cos^2\phi}=\frac y{1-y}$

Put the values of $\cot^2\phi,\cot^2\theta$ in $(1)$


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