(a) Combine the fractions: \[\frac{\sin^3\theta(1+\sin\theta)-\sin^2\theta(\sin\theta-1)}{(\sin\theta-1)(1+\sin\theta)}=\frac{\sin^4\theta+\sin^2\theta}{\sin^2\theta-1}.\] Since \(\sin^2\theta-1=-\cos^2\theta\), this is \[-\frac{\sin^2\theta}{\cos^2\theta}(1+\sin^2\theta)=-\tan^2\theta(1+\sin^2\theta).\] The identity holds where the original denominators are nonzero.

(b) Using the identity, \[-\tan^2\theta(1+\sin^2\theta)=\tan^2\theta(1-\sin^2\theta).\] Thus \(2\tan^2\theta=0\), so \(\tan\theta=0\). The only value in the open interval is \[\theta=\pi.\] It satisfies the original equation.