Can we prove this trig identity to solve this equation?

Can we prove this trig identity to solve this equation?

Question

1. Prove that $\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = \frac{2}{\sin A}$ for all values of $A$.

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1. Find the values of $x$ between $0^\circ$ and $360^\circ$ if $\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 1 + 3 \sin x.$

Solution

We will begin by writing the left-hand side as a single fraction.

$$\begin{align*} \frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} &= \frac{\sin^2 A + (1 + \cos A)^2}{(1 + \cos A)\sin A} \\ &= \frac{\sin^2 A + 1 + 2 \cos A + \cos^2 A}{(1 + \cos A)\sin A}. \end{align*}$$

For any

$A$

, we have the trigonometric identity

$$\sin^2 A + \cos^2 A = 1$$

and so we can write

$$\begin{align*} \frac{\sin^2 A + 1 + 2 \cos A + \cos^2 A}{(1 + \cos A)\sin A} &= \frac{2 + 2 \cos A}{(1 + \cos A)\sin A} \\ &= \frac{2(1 + \cos A)}{(1 + \cos A)\sin A} \\ &= \frac{2}{\sin A} \end{align*}$$

as we had hoped to show.

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 $270^\circ$.