<s>[INST] Eliminate $\theta$ from from $\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$ I am stuck with the following problem :

Eliminate $\theta$ from 
  $$\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$$ 
  to get the following result:
$$k^2(c^2-z^2)^2=c^2\left(\,(cx \tan \alpha-yz)^2+(cy \cot \alpha-zx)^2\,\right)$$

I will be grateful if someone explains the problem which was actually the part of another problem  . Thanks in advance for your time.
 [/INST] Write your line as
$$
\frac{x-k\sin\theta\cot\alpha}{k\cos\theta}=\frac{z}{c}\quad\text{and}\quad\frac{y-k\cos\theta\tan\alpha}{k\sin\theta}=\frac{z}{c}.
$$
Manipulate the two equations into
\begin{align*}
cx/k&=c\cot\alpha\sin\theta+z\cos\theta\\
cy/k&=z\sin\theta+c\tan\alpha\cos\theta
\end{align*}
so solving for $\sin\theta,\cos\theta$:
\begin{align*}
k(c^2-z^2)\sin\theta &=c\tan\alpha\cdot cx-z\cdot cy\\
k(c^2-z^2)\cos\theta &=c\cot\alpha\cdot cy-z\cdot cx
\end{align*}
Now remember $\sin^2\theta+\cos^2\theta=1$.
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