If the eccentricity of the hyperbola $x^2-y^2{\sec }^2\alpha =5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2{\sec }^2\alpha +y^2=25,\mathrm{}$ then ${\tan }^2\alpha$ is equal to

Let eccentricity of the hyperbola $\frac{x^2}{5}-\frac{y^2}{5{\cos }^2\alpha }=1$ is
$e_1^2=1+\frac{b^2}{a^2}=1+\frac{5{\cos }^2\alpha }{5}=1+{\cos }^2\alpha ;$
Similarly, eccentricity of the ellipse $\frac{x^2}{25{\cos }^2\alpha }+\frac{y^2}{25}=1$ is

$e_2^2=1-\frac{25{\cos }^2\alpha }{25}$$={\sin }^2\alpha ;$

Putting $e_1=\sqrt{3}{e}_2\Rightarrow e_1^2=3e_2^2$
$\Rightarrow 1+{\cos }^2\alpha =3{\sin }^2\alpha \Rightarrow 2=4{\sin }^2\alpha$
$\Rightarrow \sin \alpha =\frac{1}{\sqrt{2}}\Rightarrow {\tan }^2\alpha =1$