<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] 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

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

</s>