Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.

From your solution,
\begin{gather*}
x\cos \theta =y\sin \theta \\
y=x\cot \theta \\
\end{gather*}
Substitute the corresponding value of y into any of the given two equations, and you should be able to get the values of x any y both.
Let's see what we get by substituting it  into the first equation:
\begin{gather*}
2x=y\tan \theta +\sin \theta \\
2x=x+\sin \theta \\
x=\sin \theta \\
y=x\cot \theta =\cos \theta \\
\\
\end{gather*}
Can you now prove what is required?
