(i) Start with the left-hand side of the identity:

\(\csc 2\theta + \cot 2\theta = \frac{1}{\sin 2\theta} + \frac{\cos 2\theta}{\sin 2\theta}\)

Combine the terms:

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

Using the double angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\), substitute:

\(= \frac{1 + (1 - 2\sin^2 \theta)}{2\sin \theta \cos \theta}\)

\(= \frac{2 - 2\sin^2 \theta}{2\sin \theta \cos \theta}\)

\(= \frac{2(1 - \sin^2 \theta)}{2\sin \theta \cos \theta}\)

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

\(= \frac{\cos \theta}{\sin \theta}\)

\(= \cot \theta\)

Thus, the identity is proven.

(ii) Given \(\csc 2\theta + \cot 2\theta = 2\), use the proven identity:

\(\cot \theta = 2\)

\(\theta = \cot^{-1}(2)\)

Calculate \(\theta\):

\(\theta = 26.6^\circ\)

Since \(\cot \theta\) is periodic with period \(180^\circ\), the other solution is:

\(\theta = 26.6^\circ + 180^\circ = 206.6^\circ\)

Thus, \(\theta = 26.6^\circ, 206.6^\circ\).