Prove the identity $\frac{1 - \sin^4 x}{\sin^2 x} = \cot^2 x + \cos^2 x$. Hence solve the equation $\cot^2 2x + \cos^2 2x = 0$ for $0^\circ < x < 180^\circ$.

1. Split the fraction on the left-hand side into two terms.

$\text{LHS} = \frac{1}{\sin^2 x} - \frac{\sin^4 x}{\sin^2 x}$

2. Simplify the terms using reciprocal identities and exponent rules.

$\text{LHS} = \operatorname{cosec}^2 x - \sin^2 x$

3. Apply the Pythagorean identity for cosecant.

$\text{LHS} = (1 + \cot^2 x) - \sin^2 x$

4. Rearrange and apply the Pythagorean identity for sine and cosine to reach the right-hand side.

$\text{LHS} = \cot^2 x + (1 - \sin^2 x)$
$= \cot^2 x + \cos^2 x$
$= \text{RHS}$

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

From part (a), we have $\frac{1 - \sin^4 \theta}{\sin^2 \theta} = \cot^2 \theta + \cos^2 \theta$.
Substituting $\theta = 2x$, the equation $\cot^2 2x + \cos^2 2x = 0$ becomes:
$\frac{1 - \sin^4 2x}{\sin^2 2x} = 0$

2. Solve for $\sin 2x$

For the fraction to be zero, the numerator must be zero (and the denominator non-zero):
$1 - \sin^4 2x = 0$
$\sin^4 2x = 1$
$\sin^2 2x = 1$
$\sin 2x = 1 \quad \text{or} \quad \sin 2x = -1$

3. Find the values of $2x$ in the required range

Given $0^\circ < x < 180^\circ$, the range for $2x$ is $0^\circ < 2x < 360^\circ$.
For $\sin 2x = 1$:
$2x = 90^\circ$
For $\sin 2x = -1$:
$2x = 270^\circ$

4. Solve for $x$

$x = 45^\circ \quad \text{or} \quad x = 135^\circ$