Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$ Eliminate $\theta$ in following equations 
$$\begin{align}
a \cos(\theta-\alpha) &= x \\
b \cos(\theta- \beta) &=y
\end{align}$$
I am trying to solve this problem but still I am unable to get the perfect answer 
I added both the equations but it transformed it to
 $2 \cos(\theta+(\alpha + \beta)/2)$
 A: \begin{align}
  \frac{x}{a}+\frac{y}{b} &= \cos (\theta-\alpha)+\cos (\theta-\beta) \\
  &=2\cos \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta-2\theta}{2} \\
  \frac{x}{a}-\frac{y}{b} &= \cos (\theta-\alpha)-\cos (\theta-\beta) \\
  &=-2\sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta-2\theta}{2} \\
  1 &=
  \left(  
    \frac{\frac{x}{a}+\frac{y}{b}}{2\cos \frac{\alpha-\beta}{2}}
  \right)^2+
  \left(
    -\frac{\frac{x}{a}-\frac{y}{b}}{2\sin \frac{\alpha-\beta}{2}}
  \right)^2 \\
  \sin^2 (\alpha-\beta) &=
  \frac{x^2}{a^2}-\frac{2xy\cos (\alpha-\beta)}{ab}+\frac{y^2}{b^2}
\end{align}

  
*
  
*The curve is known as Lissajous figure (with same frequencies).
  
*The area bounded by the ellipse is $A=\pi ab\sin (\beta-\alpha)$.
  
*$A>0$ gives anti-clockwise trace whereas $A<0$ for clockwise.
  
*When the curve degenerates to a line segment, $A=0$


A: Hint:

 $$\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)$$

A: For simplicity, define $u:=x/a$ and $v:=y/b$, so that we have
$$\begin{align}
u &= \cos(\theta-\alpha) = \cos\theta \cos\alpha + \sin\theta\sin\alpha \\
v &= \cos(\theta-\beta) = \cos\theta\cos\beta + \sin\theta\sin\beta
\end{align}$$
This is a linear system in $\cos\theta$ and $\sin\theta$. Solving, we obtain
$$\cos\theta = \frac{v\sin\alpha - u \sin\beta}{
  \sin\alpha \cos\beta - \cos\alpha\sin\beta} = \frac{v\sin\alpha-u\sin\beta}{\sin(\alpha-\beta)} \qquad\qquad \sin\theta = \frac{u \cos\beta - v \cos\alpha}{\sin(\alpha-\beta)}$$
Then, because $\cos^2\theta + \sin^2\theta = 1$, we can write
$$\frac{\left(v\sin\alpha-u\sin\beta\right)^2}{\sin^2(\alpha-\beta)} + \frac{\left(u\cos\beta-v\cos\alpha\right)^2}{\sin^2(\alpha-\beta)} = 1$$
so that 
$$u^2 + v^2 - 2 u v (\sin\alpha\sin\beta+\cos\alpha\cos\beta) = \sin^2(\alpha-\beta)$$
which becomes

$$u^2 + v^2 - 2 u v \cos(\alpha-\beta) = \sin^2(\alpha-\beta)$$

