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.
 A: You got $x=y\tan\theta$. Now substitute back into the original equations to get $y=\cos\theta$ and $x=y\tan\theta=\sin\theta$.
A: The system
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
is a linear system of two (independent) equations in two variables, and it is readly checked that $x=\sin\theta, y=\cos\theta$ is a solution. Therefore it is the unique solution. Now $x^2+y^2=1$ follows.
A: We have from your last step $3x\cos(\theta)=3y\sin(\theta)$ thus $x=y\tan(\theta)$ putting this in (1) we get $$2y\tan(\theta)=y\tan(\theta)+\sin(\theta) \implies y=\cos(\theta)$$ from this we get $x=\sin(\theta)$ now note the identity that $\sin^2(\theta)+\cos^2(\theta)=1$
A: 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?
A: You are almost there. You already have $$2x\cos\theta=y\sin\theta+\sin\theta\cos\theta \tag1$$
and $$x\cos \theta = y \sin \theta \tag 2$$
Then $$x\cos \theta = y \sin \theta = 2x\cos \theta - y \sin \theta = \sin \theta \cos \theta$$
and you know $\sin \theta \ne 0, \cos \theta \ne 0$. Can you end it now?
