Review question

# Can we prove this trig identity to solve this equation? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5269

## Solution

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

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

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 $270^\circ$.