Showing that $\frac{x^2+2x\cos2\alpha+1}{x^2+2x\cos2\beta+1}$ lies between $\frac{\cos^2\alpha}{\cos^2\beta}$ and $\frac{\sin^2\alpha}{\sin^2\beta}$

Question

If $β$ is such that $\sin\beta≠0$, then show that the expression $$\frac{x^2 + 2x\cos2\alpha + 1}{x^2 + 2x\cos2\beta+1}$$ always lies between $\dfrac{\cos^2\alpha}{\cos^2\beta}$ and $\dfrac{\sin^2\alpha}{\sin^2\beta}$.

I tried taking the whole expression as $Y$ and solved until a point I got $$(\cos2\alpha - \cos2\beta)^2 \geq 0$$ That, however, does not give me the answer. Can anyone point out where I went wrong?

Answer 1

Method$\#1:$

Let the expression be equal to $k$

Rearrange to form a quadratic equation in $x$

As $x$ is real, the discriminant must be $\ge0$

Method $\#2:$

Let the given expression $=y$

Find $\dfrac1{y-1}$

Divide numerator & denominator by $x$

Now for real $x>0$ $$x+\dfrac1x\ge2$$

and what happens if $x<0$