(i) Start with the left-hand side (LHS):

\(\tan x + \frac{1}{\tan x} = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\)

Combine the fractions:

\(= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\)

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

\(= \frac{1}{\sin x \cos x}\)

Thus, the identity is proven.

(ii) Given equation:

\(\frac{2}{\sin x \cos x} = 1 + 3 \tan x\)

Using part (i), substitute \(\frac{2}{\sin x \cos x} = 2 \left( \tan x + \frac{1}{\tan x} \right)\):

\(2 \left( \tan x + \frac{1}{\tan x} \right) = 3 \tan x + 1\)

Multiply through by \(\tan x\):

\(2 \tan^2 x + 2 = 3 \tan x + \tan x\)

Rearrange to form a quadratic equation:

\(2 \tan^2 x + \tan x - 2 = 0\)

Solving the quadratic equation, we find:

\(\tan x = 1\) or \(\tan x = -2\)

For \(\tan x = 1\), \(x = 45^\circ\).

For \(\tan x = -2\), \(x = 116.6^\circ\).