(a) Wherever both sides are defined, combine the fractions: \[\begin{aligned}\frac{\cos\theta}{1-\sin\theta}-\frac1{\cos\theta}&=\frac{\cos^2\theta-(1-\sin\theta)}{(1-\sin\theta)\cos\theta}\\&=\frac{1-\sin^2\theta-1+\sin\theta}{(1-\sin\theta)\cos\theta}\\&=\frac{\sin\theta(1-\sin\theta)}{(1-\sin\theta)\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta.\end{aligned}\]

(b) Using (a), \(2\tan\theta+3=0\), so \(\tan\theta=-3/2\). The reference angle is \(\tan^{-1}(3/2)=56.3099\ldots^\circ\). Tangent is negative in quadrants II and IV: \[\boxed{\theta=123.7^\circ,303.7^\circ}\quad(1\text{ d.p.}).\] Both satisfy the original denominator restrictions.