- Prove that \[ \frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = \frac{2}{\sin A} \] for all values of \(A\).

as we had hoped to show.

- Find the values of \(x\) between \(0^\circ\) and \(360^\circ\) if \[ \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 1 + 3 \sin x. \]

From the previous part of the question, \(x\) satisfies \[ \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 1 + 3 \sin x. \] exactly when \[ \frac{2}{\sin x} = 1 + 3 \sin x. \] Since this never holds when \(\sin x = 0\), we can multiply through by \(\sin x\) and rearrange to produce \[ 3 \sin^2 x + \sin x - 2 = 0. \] This is a quadratic equation in \(\sin x\); we can factorise it as \[ (3 \sin x - 2)(\sin x + 1) = 0. \] Thus \(x\) satisfies the original equation given in the question exactly when \[ \sin x = -1 \quad\text{or}\quad \sin x = \frac{2}{3}. \]

The unique solution to the first of these possibilities with \(0^\circ \le x \le 360^\circ\) is \(x = 270^\circ\).

The second possibility gives two solutions with \(0^\circ \le x \le 360^\circ\).

We have \(\sin^{-1} \left( \dfrac{2}{3} \right)\), which lies between \(0^\circ\) and \(90^\circ\), and \(180^\circ - \sin^{-1} \left( \dfrac{2}{3} \right)\), which lies between \(90^\circ\) and \(180^\circ\).

So there are three possible solutions in total, \(41.8^\circ, 138.2^\circ\), and \(270^\circ\).