First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = 2 + 2\cos 2\theta\)

\(\frac{dy}{d\theta} = 2\sin 2\theta\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{2\sin 2\theta}{2 + 2\cos 2\theta}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{2\sin 2\theta}{2(1 + \cos 2\theta)} = \frac{\sin 2\theta}{1 + \cos 2\theta}\)

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 + \cos 2\theta = 2\cos^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta\)

(a) Differentiate \(x = te^{2t}\) with respect to \(t\):

\(\frac{dx}{dt} = e^{2t} + 2te^{2t}\).

Differentiate \(y = t^2 + t + 3\) with respect to \(t\):

\(\frac{dy}{dt} = 2t + 1\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 1}{e^{2t}(1 + 2t)}\).

Simplify to obtain \(\frac{dy}{dx} = e^{-2t}\).

(b) At \(t = -1\), find \(x\) and \(y\):

\(x = -\frac{1}{e^2}\), \(y = 3\).

The gradient of the tangent is \(e^{-2(-1)} = e^2\), so the gradient of the normal is \(-\frac{1}{e^2}\).

The equation of the normal is \(y - 3 = -\frac{1}{e^2}(x + \frac{1}{e^2})\).

Substitute \(x = 0\) to find \(y\):

\(y = 3 - \frac{1}{e^4}\).

Thus, the normal passes through \(\left( 0, 3 - \frac{1}{e^4} \right)\).

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

\(\\frac{dx}{dt} = \\frac{d}{dt}(2t - \\tan t) = 2 - \\sec^2 t\).

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

\(\\frac{dy}{dt} = \\frac{d}{dt}(\\ln(\\sin 2t)) = \\frac{1}{\\sin 2t} \\cdot \\frac{d}{dt}(\\sin 2t) = \\frac{1}{\\sin 2t} \\cdot 2 \\cos 2t = \\frac{2 \\cos 2t}{\\sin 2t}\).

Now, use the chain rule to find \(\\frac{dy}{dx}\):

\(\\frac{dy}{dx} = \\frac{dy}{dt} \\div \\frac{dx}{dt} = \\frac{2 \\cos 2t}{\\sin 2t} \\div (2 - \\sec^2 t)\).

Simplify using trigonometric identities:

\(\\frac{dy}{dx} = \\frac{2 \\cos 2t}{\\sin 2t} \\cdot \\frac{1}{2 - \\sec^2 t}\).

Using the double angle formula, \(\\sin 2t = 2 \\sin t \\cos t\), and \(\\cos 2t = \\cos^2 t - \\sin^2 t\), we simplify:

\(\\frac{dy}{dx} = \\frac{\\cos t}{\\sin t} = \\cot t\).

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

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = \\sin \theta\)

\(\frac{dy}{d\theta} = -\\sin \theta + \frac{1}{2} \\sin 2\theta\)

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-\\sin \theta + \frac{1}{2} \\sin 2\theta}{\\sin \theta}\)

Simplify using the double angle identity \(\sin 2\theta = 2\\sin \theta \\cos \theta\):

\(\frac{dy}{dx} = \frac{-\\sin \theta + \\sin \theta \\cos \theta}{\\sin \theta} = -1 + \\cos \theta\)

Use the identity \(\\cos \theta = 1 - 2\\sin^2 \left( \frac{1}{2} \theta \right)\):

\(\frac{dy}{dx} = -1 + 1 - 2\\sin^2 \left( \frac{1}{2} \theta \right) = -2\\sin^2 \left( \frac{1}{2} \theta \right)\)