# 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 $\frac{\sin^2\alpha}{\sin^2\beta}$

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?

• Welcome to the StackExchange community. Please include your attempt so, that we can understand from where you went wrong. Also, Please avoid putting the question in the title. – Kumar Jun 21 '19 at 1:44

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 $$x<0$$