Prove that, if \(t=\tan \tfrac{1}{2}\theta\), then \[\sin\theta=\frac{2t}{1+t^2}\quad\text{and}\quad \cos\theta=\frac{1-t^2}{1+t^2}.\]

By expressing \(\dfrac{3+\cos\theta}{\sin\theta}\) in terms of \(t\), or otherwise, show that this expression cannot have any value between \(-2\sqrt{2}\) and \(+2\sqrt{2}\).