Prove the identity $\dfrac{\tan A - \cot A}{\tan A + \cot A} = 1 - 2\cos^{2} A$. Hence solve the equation $\frac{\tan 2\theta - \cot 2\theta}{\tan 2\theta + \cot 2\theta} = \frac{1}{2}$ for $0 < \theta < \pi$, giving your answers in terms of $\pi$.

1. Express the left-hand side (LHS) in terms of sine and cosine.

$\text{LHS} = \frac{\tan A - \cot A}{\tan A + \cot A} = \frac{\frac{\sin A}{\cos A} - \frac{\cos A}{\sin A}}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

2. Simplify the complex fraction by multiplying the numerator and denominator by $\sin A \cos A$.

$\text{LHS} = \frac{\sin^2 A - \cos^2 A}{\sin^2 A + \cos^2 A}$

3. Apply the Pythagorean identity to the denominator.

Since $\sin^2 A + \cos^2 A = 1$,
$\text{LHS} = \frac{\sin^2 A - \cos^2 A}{1} = \sin^2 A - \cos^2 A$

4. Substitute $\sin^2 A = 1 - \cos^2 A$ to obtain the right-hand side (RHS).

$\text{LHS} = (1 - \cos^2 A) - \cos^2 A$
$\text{LHS} = 1 - 2\cos^2 A = \text{RHS}$

1. Apply the identity from part (a) to the given equation.

From part (a), $\frac{\tan A - \cot A}{\tan A + \cot A} = 1 - 2\cos^{2} A$.
Substituting $A = 2\theta$, the equation becomes:
$1 - 2\cos^{2}(2\theta) = \frac{1}{2}$

2. Rearrange to solve for $\cos^{2}(2\theta)$ and then $\cos(2\theta)$.

$2\cos^{2}(2\theta) = 1 - \frac{1}{2} = \frac{1}{2}$
$\cos^{2}(2\theta) = \frac{1}{4}$
$\cos(2\theta) = \pm \frac{1}{2}$

3. Determine the range for $2\theta$ and find the basic angle.

Given $0 < \theta < \pi$, the range for $2\theta$ is $0 < 2\theta < 2\pi$.
The basic angle $\alpha$ for $\cos \alpha = \frac{1}{2}$ is $\frac{\pi}{3}$.

4. Find all possible values for $2\theta$ and then $\theta$.

For $\cos(2\theta) = \frac{1}{2}$:
$2\theta = \frac{\pi}{3}, 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
For $\cos(2\theta) = -\frac{1}{2}$:
$2\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}, \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
So, $2\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
Dividing by 2:
$\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}$