证明恒等式 $\frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta} = 2\csc \theta$。 由此，求在 $0^\circ \le \theta \le 360^\circ$ 范围内满足方程
$\frac{1+\cos 2\theta}{\sin 2\theta} + \frac{\sin 2\theta}{1+\cos 2\theta} = \tan 75^\circ.$
的所有角度。

1. Combine the fractions on the left-hand side into a single fraction

$\text{LHS} = \frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta}$
$= \frac{(1+\cos \theta)^2 + \sin^2 \theta}{\sin \theta (1+\cos \theta)}$

2. Expand the numerator

$= \frac{1 + 2\cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta (1+\cos \theta)}$

3. Apply the Pythagorean identity to simplify the numerator

Since $\cos^2 \theta + \sin^2 \theta = 1$,
$= \frac{1 + 2\cos \theta + 1}{\sin \theta (1+\cos \theta)}$
$= \frac{2 + 2\cos \theta}{\sin \theta (1+\cos \theta)}$

4. Factorise the numerator and simplify the fraction to reach the right-hand side

$= \frac{2(1 + \cos \theta)}{\sin \theta (1+\cos \theta)}$
$= \frac{2}{\sin \theta}$
$= 2\csc \theta = \text{RHS}$

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

From (a), $\frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta} = 2\csc \theta$.
Substituting $\theta$ with $2\theta$, the equation becomes:
$2\csc 2\theta = \tan 75^\circ$

2. Rearrange to find $\sin 2\theta$

$\frac{2}{\sin 2\theta} = \tan 75^\circ$
$\sin 2\theta = \frac{2}{\tan 75^\circ}$
$\sin 2\theta = 0.535898$

3. Find the basic angle and solve for $2\theta$ in the required range

Since $0^\circ \le \theta \le 360^\circ$, we have $0^\circ \le 2\theta \le 720^\circ$.
Basic angle $\alpha = \sin^{-1}(0.535898) \approx 32.405^\circ$
Since $\sin 2\theta > 0$, $2\theta$ is in the 1st and 2nd quadrants.
$2\theta = 32.405^\circ, 180^\circ - 32.405^\circ, 360^\circ + 32.405^\circ, 540^\circ - 32.405^\circ$
$2\theta = 32.405^\circ, 147.595^\circ, 392.405^\circ, 507.595^\circ$

4. Solve for $\theta$ and round to 1 decimal place

$\theta = 16.2025^\circ, 73.7975^\circ, 196.2025^\circ, 253.7975^\circ$
$\theta = 16.2^\circ, 73.8^\circ, 196.2^\circ, 253.8^\circ \text{ (to 1 d.p.)}$