<s>[INST] 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
 $2 \cos(\theta+(\alpha + \beta)/2)$
 [/INST] \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, $A=0$


</s>