Is there an alternate method to prove the following?
If $2\cos A=x + \frac{1}{x}$, $2\cos B=y+\frac{1}{y}$ then show that $2\cos(A-B) = \frac{x}{y} + \frac{y}{x}$
I know that it can be done by converting the first two equations into quadratic equations and then calculating $\frac{x}{y}$ and $\frac{y}{x}$ and then adding them, but are there any simpler methods?
Hint:
$$x^2-2x\cos A+1=0\implies x=\cos A\pm i\sin A$$
Similarly, $$y=\cos B\pm i\sin B$$
Case$\#1:$
$$ x=\cos A+i\sin A,y=\cos B+i\sin B$$
Case$\#2:$
$$ x=\cos A+i\sin A,y=\cos B-+i\sin B$$
Case$\#3:$
$$ x=\cos A-i\sin A,y=\cos B+i\sin B$$
Case$\#4:$
$$ x=\cos A-i\sin A,y=\cos B-i\sin B$$
If you don't mind working with complex numbers (although $x$ and $y$ are both complex anyway), you can use the fact that $$ 2\cos \theta = e^{i\theta} + e^{-i\theta} $$
and conclude that $x = e^{iA}$, $y = e^{iB}$. Then $$ 2\cos(A-B) = e^{i(A-B)} + e^{-i(A-B)} = \frac{x}{y} + \frac{y}{x} $$
Technically, there are 4 multiple possibilities depending on how you pick the sign $$ x = e^{\pm iA}, \ y = e^{\pm iB} $$
An alternate result would be if $x =e^{iA}$, $y = e^{-iB}$. Then $$ 2\cos(A-B) = xy + \frac{1}{xy} $$
Another approach without using Euler's identity is to find $$ 4\sin^2 A = 4 - \cos^2 A = 4 - \left(x + \frac{1}{x}\right)^2 = 2 - x^2 - \frac{1}{x^2} = -\left(x - \frac{1}{x}\right)^2 $$
Therefore $$ 2\sin A = \pm i\left(x - \frac{1}{x}\right), \ 2\sin B = \pm i\left(y - \frac{1}{y}\right) $$
Then apply the difference of angles formula $$ \begin{align} 4\cos(A-B) &= 4\cos A\cos B + 4\sin A\sin B \\ &= \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right) \pm \left(x - \frac{1}{x}\right)\left(y - \frac{1}{y}\right) \end{align} $$
Hint (assuming reals): if $\,x \gt 0\,$ then $\,x+ \dfrac{1}{x} \ge 2\,$, else if $\,x \lt 0\,$ then $\,x+ \dfrac{1}{x} \le -2\,$.
If you stick to the reals, using @dxiv's remark, $x=\pm1,y=\pm1$ so that $A=k\pi,B=k'\pi$ and $A-B=k''\pi$, so that $2\cos(A-B)=\pm2=\dfrac xy+\dfrac yx$.