Prove the identity $\csc 4x + \cot 4x = \cot 2x$. Hence, find, for $-180^\circ \le \theta \le 180^\circ$, the values of $\theta$ for which $\csc 2\theta + \cot 2\theta = -\sqrt{3}$.

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

$\text{LHS} = \csc 4x + \cot 4x$
$= \frac{1}{\sin 4x} + \frac{\cos 4x}{\sin 4x}$
$= \frac{1 + \cos 4x}{\sin 4x}$

2. Apply double angle identities to the numerator and denominator

Using $\cos 2A = 2\cos^2 A - 1$ and $\sin 2A = 2\sin A \cos A$ with $A = 2x$:
$= \frac{1 + (2\cos^2 2x - 1)}{2\sin 2x \cos 2x}$
$= \frac{2\cos^2 2x}{2\sin 2x \cos 2x}$

3. Simplify the fraction to obtain the right-hand side

$= \frac{\cos 2x}{\sin 2x}$
$= \cot 2x$
$= \text{RHS}$

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

From part (a), we have $\csc 4x + \cot 4x = \cot 2x$.
Let $4x = 2\theta$, which means $2x = \theta$.
Substituting this into the identity gives:
$\csc 2\theta + \cot 2\theta = \cot \theta$

2. Rewrite the equation in terms of a single trigonometric function

The equation $\csc 2\theta + \cot 2\theta = -\sqrt{3}$ becomes:
$\cot \theta = -\sqrt{3}$
$\tan \theta = -\frac{1}{\sqrt{3}}$

3. Find the basic angle

Basic angle $\alpha = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$

4. Find the values of $\theta$ in the given range

Since $\tan \theta$ is negative, $\theta$ lies in the 2nd and 4th quadrants.
For $-180^\circ \le \theta \le 180^\circ$:
In the 2nd quadrant: $\theta = 180^\circ - 30^\circ = 150^\circ$
In the 4th quadrant (negative angle): $\theta = -30^\circ$