证明恒等式 $\frac{\tan^2 y - \sin^2 y}{\sec^2 y (\csc^2 y - 1)} = \tan^2 y \sin^4 y$。 由此求解方程 $\frac{\tan^2 y - \sin^2 y}{\sec^2 y (\csc^2 y - 1)} = \frac{1}{16} \tan^2 y$，其中 $0^\circ \le y \le 180^\circ$。

1. Simplify the numerator by expressing $\tan^2 y$ in terms of sine and cosine, and factorising.

$\text{Numerator} = \tan^2 y - \sin^2 y = \frac{\sin^2 y}{\cos^2 y} - \sin^2 y = \sin^2 y \left( \frac{1}{\cos^2 y} - 1 \right)$

2. Apply Pythagorean identities to further simplify the numerator.

$\sin^2 y \left( \sec^2 y - 1 \right) = \sin^2 y \tan^2 y$

3. Simplify the denominator using Pythagorean and quotient identities.

$\text{Denominator} = \sec^2 y (\csc^2 y - 1) = \sec^2 y \cot^2 y = \frac{1}{\cos^2 y} \cdot \frac{\cos^2 y}{\sin^2 y} = \frac{1}{\sin^2 y}$

4. Combine the simplified numerator and denominator to prove the identity.

$\text{LHS} = \frac{\sin^2 y \tan^2 y}{\frac{1}{\sin^2 y}} = \sin^2 y \tan^2 y \cdot \sin^2 y = \tan^2 y \sin^4 y = \text{RHS}$

1. Substitute the identity from part (i) into the equation

Using the result from (i), the equation becomes:
$\tan^2 y \sin^4 y = \frac{1}{16} \tan^2 y$

2. Rearrange and factorise the equation

$\tan^2 y \sin^4 y - \frac{1}{16} \tan^2 y = 0$
$\tan^2 y \left( \sin^4 y - \frac{1}{16} \right) = 0$

3. Solve for $\tan y = 0$ and check validity

$\tan^2 y = 0 \implies \tan y = 0$
$y = 0^\circ, 180^\circ$
However, at $y = 0^\circ$ and $y = 180^\circ$, $\csc y$ is undefined, so these solutions are rejected.

4. Solve for $\sin^4 y = \frac{1}{16}$

$\sin^4 y = \frac{1}{16}$
$\sin y = \pm \frac{1}{2}$
Since $0^\circ \le y \le 180^\circ$, $\sin y \ge 0$, so:
$\sin y = \frac{1}{2}$
$y = 30^\circ, 150^\circ$