To find the value of 'a' that satisfies the given condition, we need to compare the eccentricities of the hyperbola and the ellipse.

1. Equation of the hyperbola: x^2 - y^2sec^2(a) = 5
2. Equation of the ellipse: x^2sec^2(a) + y^2 = 25

Let's find the eccentricities of both the hyperbola and the ellipse.

Eccentricity of the hyperbola:
The eccentricity of a hyperbola is given by the formula e = √(1 + b^2/a^2), where a and b are the semi-major and semi-minor axes, respectively.

From the equation of the hyperbola, we can observe that a = √5 and b = √(5sec^2(a)). Substituting these values into the eccentricity formula, we get:
e_h = √(1 + (5sec^2(a))/(5)) = √(1 + sec^2(a)) = √(1 + 1/cos^2(a))

Eccentricity of the ellipse:
The eccentricity of an ellipse is given by the formula e = √(1 - b^2/a^2), where a and b are the semi-major and semi-minor axes, respectively.

From the equation of the ellipse, we can observe that a = √25 = 5 and b = √(25sec^2(a)) = 5sec(a). Substituting these values into the eccentricity formula, we get:
e_e = √(1 - (25sec^2(a))/(25)) = √(1 - sec^2(a)) = √(1 - 1/cos^2(a))

Now, we are given that the eccentricity of the hyperbola is times the eccentricity of the ellipse. Mathematically, this can be written as:
e_h = k * e_e

Substituting the expressions for e_h and e_e derived earlier, we have:
√(1 + 1/cos^2(a)) = k * √(1 - 1/cos^2(a))

Squaring both sides of the equation, we get:
1 + 1/cos^2(a) = k^2 * (1 - 1/cos^2(a))

Simplifying the equation, we have:
1 = k^2 - k^2/cos^2(a)
1 = k^2(1 - 1/cos^2(a))

Since cos^2(a) can never be negative, we can divide both sides of the equation by (1 - 1/cos^2(a)), which gives:
1/(1 - 1/cos^2(a)) = k^2

Simplifying further, we get:
cos^2(a)/(cos^2(a) - 1) = k^2

Using the identity cos^2(a) - sin^2(a) = 1, we can rewrite the equation as:
cos^2(a)/(sin^2(a)) = k^2
cot^2(a) = k^2

Taking the square root of both sides, we get:
cot(a) = k

We know that cot(a) = 1/tan(a). Sub