\[ \text { Find the value of } \cos ^2 \theta\left(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\right) \]

Given:
– We need to find the value of \(\cos^2 \theta \left(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} + \sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\right)\).

1. Multiply the numerator and denominator of each fraction inside the square roots by the conjugate of the denominator:

\[ \cos^2 \theta \left(\sqrt{\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}} + \sqrt{\frac{(1-\sin \theta)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}}\right) \]

2. Simplify the expressions inside the square roots using the identity \(1 – \sin^2 \theta = \cos^2 \theta\):

\[ \cos^2 \theta \left(\sqrt{\frac{(1+\sin \theta)^2}{\cos^2 \theta}} + \sqrt{\frac{(1-\sin \theta)^2}{\cos^2 \theta}}\right) \]

3. Simplify further by taking the square roots:

\[ \cos^2 \theta \left(\frac{1+\sin \theta}{\cos \theta} + \frac{1-\sin \theta}{\cos \theta}\right) \]

4. Combine the fractions:

\[ \cos^2 \theta \left(\frac{1+\sin \theta + 1 – \sin \theta}{\cos \theta}\right) = \frac{2\cos^2 \theta}{\cos \theta} \]

5. Simplify the expression:

\[ 2\cos \theta \]

Conclusion:
The value of \(\cos^2 \theta \left(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} + \sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\right)\) is \(2\cos \theta\).