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

 For the hyperbola

$\frac{{\mathrm{x}}^2}{5}-\frac{{\mathrm{y}}^2}{5{\cos }^2\mathrm{\alpha }}=1$

we have 

 ${\mathrm{e}}_1^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=1+\frac{5{\cos }^2\mathrm{\alpha }}{5}=1+{\cos }^2\mathrm{\alpha }$

For the ellipse

  $\frac{{\mathrm{x}}^2}{25{\cos }^2\mathrm{\alpha }}+\frac{{\mathrm{y}}^2}{25}=1$

we have

  ${\mathrm{e}}_2^2=1-\frac{25{\cos }^2\mathrm{\alpha }}{25}={\sin }^2\mathrm{\alpha }$

Given that

$\begin{array}{l}{\mathrm{e}}_1=\sqrt{3}{\mathrm{e}}_2\\ \therefore {\mathrm{e}}_1^2=3{\mathrm{e}}_2^2\\ \Rightarrow 1+{\cos }^2\mathrm{\alpha }=3{\sin }^2\mathrm{\alpha }\\ \Rightarrow 2=4{\sin }^2\mathrm{\alpha }\\ \Rightarrow \sin \mathrm{\alpha }=\frac{1}{\sqrt{2}}\end{array}$