(a) To find \(\frac{dy}{dx}\), we use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\).

First, find \(\frac{dx}{dt}\):

\(x = \frac{1}{\cos t} \Rightarrow \frac{dx}{dt} = \sec t \tan t\).

Next, find \(\frac{dy}{dt}\):

\(y = \ln \tan t \Rightarrow \frac{dy}{dt} = \frac{1}{\tan t} \cdot \sec^2 t = \frac{\sec^2 t}{\tan t}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{\sec^2 t}{\tan t}}{\sec t \tan t} = \frac{\sec t}{\tan^2 t}\).

Since \(\tan t = \frac{\sin t}{\cos t}\), we have \(\tan^2 t = \frac{\sin^2 t}{\cos^2 t}\).

Thus, \(\frac{dy}{dx} = \frac{\sec t \cos^2 t}{\sin^2 t} = \frac{\cos t}{\sin^2 t}\).

(b) When \(y = 0\), \(\ln \tan t = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4}\).

At \(t = \frac{\pi}{4}\), \(x = \frac{1}{\cos \frac{\pi}{4}} = \sqrt{2}\).

The gradient \(\frac{dy}{dx} = \frac{\cos t}{\sin^2 t} = \frac{\cos \frac{\pi}{4}}{\sin^2 \frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{2}\).

The equation of the tangent is \(y - 0 = \sqrt{2}(x - \sqrt{2})\).

Simplifying gives \(y = \sqrt{2}x - 2\).