If the eccentricity of the hyperbola x2−y2sec2⁡α=5 is 3 times the eccentricity of the ellipse x2sec2⁡α+y2=25,  then a value of α is

For the hyperbola x25−y25cos2⁡α=1we havee12=1+b2a2=1+5cos2⁡α5=1+cos2⁡αFor the ellipse x225cos2⁡α+y225=1we havee22=1−25cos2⁡α25=sin2⁡αGiven that    e1=3e2∴     e12=3e22 or     1+cos2⁡α=3sin2⁡α or     2=4sin2⁡α or      sin⁡α=12