If the eccentricity of the hyperbola x^ 2 - y^ 2 ^ 2 α = 5 is 3...

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

${e}_{2}^{2} = 1 - \dfrac{25\cos^{2} \alpha}{25}= \sin^{2} \alpha;$

Putting ${e}_{1} = \sqrt{3}{e}_{2} ⇒ {e}_{1}^{2} = 3{e}_{2}^{2}$
$⇒ 1 + \cos^{2} \alpha = 3\sin^{2} \alpha ⇒ 2 = 4\sin^{2} \alpha$
$⇒ sin \alpha = \dfrac{1}{\sqrt{2}} ⇒ \tan^{2} \alpha = 1$