A trigonemetry problem

Question

If $\sec(\theta + \alpha)$$\sec(\theta-\alpha)$ = $2\sec(\theta)$
and $\cos(\theta)$=K $\cos{\frac \alpha2}$
Then prove that $K=\pm \sqrt2$
I have converted the $\sec$ into $\cos$ and then applied the formulaes of $\cos(\theta+\alpha)$ $\cos(\theta+\alpha)$ we get
$2\sin^2\alpha$ =$2\cos^2\theta$-$\cos\theta$
Then $2\cos^2(\theta)$=$K^2 (1+\cos\alpha)$ from the other equation
But i can't get $K$ by solving both the equations

Answer 1

First solve $\sec(\theta + \alpha)+\sec(\theta-\alpha)=2\sec(\theta)$

and then compare it to $\cos(\theta)=K\cos{\frac \alpha2}$

We have $$\sec(\theta+\alpha)+\sec(\theta-\alpha)=2\sec(\theta)$$ $$\frac{1}{\cos(\theta+\alpha)}+\frac{1}{\cos(\theta-\alpha)}=\frac{2}{\cos\theta}$$ $$\frac{\cos(\theta+\alpha)+\cos(\theta-\alpha)}{\cos(\theta+\alpha)\cos(\theta-\alpha)}=\frac{2}{\cos(\theta)}$$ $$\frac{2\cos\theta\cos\alpha}{\cos2\theta+\cos2\alpha}=\frac{1}{\cos\theta}$$ $$2\cos^2\theta\cos\alpha=\cos2\theta+\cos2\alpha$$ $$2\cos^2\theta(\cos\alpha-1)=2\cos^2\alpha-2$$ $$\cos^2\theta=\frac{\cos^2\alpha-1}{\cos\alpha-1}=\cos\alpha+1$$ $$\cos^2\theta=2\cos^2\left(\frac{\alpha}{2}\right)$$ $$\cos\theta=\pm \sqrt{2}\cos\left(\frac{\alpha}{2}\right)$$

When we compare this to $\cos(\theta)=K\cos{\left(\frac{\alpha}{2}\right)}$ we get $$K=\pm\sqrt{2}$$