If $\mathrm{sec\alpha }$ is  the average of $\sec (\mathrm{\alpha }-2\mathrm{\beta })\text{ and }\sec (\mathrm{\alpha }+2\mathrm{\beta })$ then the value of $\left(2{\sin }^2\mathrm{\beta }-{\sin }^2\mathrm{\alpha }\right)$ is

$\begin{array}{l}\sec \mathrm{\alpha }=\frac{\sec (\mathrm{\alpha }-2\mathrm{\beta })+\sec (\mathrm{\alpha }+2\mathrm{\beta })}{2}\\ \Rightarrow \frac{2}{\cos \mathrm{\alpha }}=\frac{\cos (\mathrm{\alpha }+2\mathrm{\beta })+\cos (\mathrm{\alpha }-2\mathrm{\beta })}{\cos (\mathrm{\alpha }-2\mathrm{\beta })\cos (\mathrm{\alpha }+2\mathrm{\beta })}\end{array}$

$\begin{array}{l}\Rightarrow \cos 2\mathrm{\alpha }+\cos 4\mathrm{\beta }=\cos \mathrm{\alpha }(2\cos \mathrm{\alpha }\cdot \cos 2\mathrm{\beta })\\ \Rightarrow 2{\cos }^2\mathrm{\alpha }-1+2{\cos }^{2}2\mathrm{\beta }-1=2{\cos }^2\mathrm{\alpha cos}2\mathrm{\beta }\\ \Rightarrow {\cos }^2\mathrm{\alpha }(1-\cos 2\mathrm{\beta })+(\cos 2\mathrm{\beta }+1)(\cos 2\mathrm{\beta }-1)=0\\ \Rightarrow (1-\cos 2\mathrm{\beta })\left({\cos }^2\mathrm{\alpha }-\cos 2\mathrm{\beta }-1\right)=0\\ \Rightarrow {\cos }^2\mathrm{\alpha }=\cos 2\mathrm{\beta }+1(\text{ as }\mathrm{\beta }\ne \mathrm{n\pi })\\ \Rightarrow {\cos }^2\mathrm{\alpha }=2{\cos }^2\mathrm{\beta }\\ \Rightarrow 1-{\sin }^2\mathrm{\alpha }=2\left(1-{\sin }^2\mathrm{\beta }\right)\\ \Rightarrow 2{\sin }^2\mathrm{\beta }-{\sin }^2\mathrm{\alpha }=1\end{array}$