Answer only one of the following two alternatives.
**EITHER**
It is given that \(w = \cos y\) and
$$\tan y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 2\tan y \frac{dy}{dx} = 1 + e^{-2x} \sec y.$$
- Show that
$$\frac{d^2 w}{dx^2} + 2\frac{dw}{dx} + w = -e^{-2x}.$$ [4]
- Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0\), \(y = \frac{1}{4}\pi\) and \(\frac{dy}{dx} = \frac{1}{\sqrt{3}}\). [10]
**OR**
The curves \(C_1\) and \(C_2\) have polar equations, for \(0 \leq \theta \leq \frac{1}{2}\pi\), as follows:
\begin{align}
C_1: r &= 2(e^\theta + e^{-\theta}),
C_2: r &= e^{2\theta} - e^{-2\theta}.
\end{align}
The curves intersect at the point \(P\) where \(\theta = \alpha\).
- Show that \(e^{2\alpha} - 2e^\alpha - 1 = 0\). Hence find the exact value of \(\alpha\) and show that the value of \(r\) at \(P\) is \(4\sqrt{2}\). [6]
- Sketch \(C_1\) and \(C_2\) on the same diagram. [3]
- Find the area of the region enclosed by \(C_1\), \(C_2\) and the initial line, giving your answer correct to 3 significant figures. [5]