(a) Start with the left-hand side:

\( \frac{1-\sin\theta}{\cos\theta}+\frac{\cos\theta}{1-\sin\theta} \).

Use a common denominator:

\( \frac{(1-\sin\theta)^2+\cos^2\theta}{\cos\theta(1-\sin\theta)} \).

Expand the numerator:

\( (1-2\sin\theta+\sin^2\theta)+\cos^2\theta \).

Since \( \sin^2\theta+\cos^2\theta=1 \), the numerator becomes

\( 2-2\sin\theta=2(1-\sin\theta) \).

Therefore

\( \frac{2(1-\sin\theta)}{\cos\theta(1-\sin\theta)}=\frac2{\cos\theta} \).

Hence the identity is proved.

(b) Using part (a),

\( \frac2{\cos\theta}=\frac{\tan^3\theta}{\sin\theta} \).

Now

\( \frac{\tan^3\theta}{\sin\theta}=\frac{\sin^2\theta}{\cos^3\theta} \).

So

\( \frac2{\cos\theta}=\frac{\sin^2\theta}{\cos^3\theta} \).

Multiply by \( \cos^3\theta \):

\( 2\cos^2\theta=\sin^2\theta \).

Using \( \sin^2\theta=1-\cos^2\theta \):

\( 2\cos^2\theta=1-\cos^2\theta \).

\( 3\cos^2\theta=1 \), so \( \cos\theta=\pm\frac1{\sqrt3} \).

Answer: \( \theta=54.7^\circ,\ 125.3^\circ,\ 234.7^\circ,\ 305.3^\circ \).