Solving equations with $\sec(x)$ , $\cosec(x)$ and $\cot(x)$

In a nutshell

Using these functions can involve simplifying expressions as well as solving equations they're involved in.

Example 1

Simplify the following expression: $\cos(x)\cot(x)\cosec(x)$ .

It may help to express everything in terms of $\sin(x)$ and $\cos(x)$ :

$\begin{aligned}\cos(x)\cot(x)\cosec(x)&=\cos(x)\left(\frac{\cos(x)}{\sin(x)}\right)\left(\frac1{\sin(x)}\right)\\&=\frac{\cos^2(x)}{\sin^2(x)}\\&=\underline{\cot^2(x)}\end{aligned}$ 

You may sometimes be able to spot a simplification without first converting everything into sines and cosines. However, if in doubt, use this method.

Solving equations

Often it will help to first simplify an equation before trying to solve it. Solving equations involving $\cosec(x)$ , $\sec(x)$ and $\cot(x)$ can be done by also expressing in terms of $\sin(x)$ , $\cos(x)$ and $\tan(x)$ where possible, since you are already familiar with these functions.

Example 2

Solve $\cosec(x)=2$ in the interval $0\leq x\leq360^\circ$ .

Reexpress in terms of $\sin(x)$ and solve:

$\begin{aligned}\cosec(x)&=2\\\frac1{\sin(x)}&=2\\\sin(x)&=0.5\\x&=\sin^{-1}(0.5)\\&=30\degree \end{aligned}$ 

Thus the principal solution is $30^\circ$ . Since you are solving a sine equation, the other solution is found by subtracting from $180$ :

$\begin{aligned}x&=180-30\\&=150\degree \end{aligned}$ 

So the solutions to $\cosec(x)=2$ in the interval $0\leq x\leq360^\circ$ are $\underline{x=30\degree}$ and $\underline{x=150\degree}$ . Note: The graph of $y=\cosec(x)$ can help to check that your solutions are in the right region. But since you turned this into a sine problem, using the cosecant graph while solving is not necessarily useful.

Maths; Trigonometric functions; KS5 Year 13; Solving equations with sec x, cosec x and cot x

Proofs

Since you now have many trigonometric functions that are closely related, you can show that some expressions are equal to others, even if they look very different.

Example 3

Prove that $\frac{\sin(\theta)}{1+\cot(\theta)}\equiv\frac{\sec(\theta)-\cos(\theta)}{\tan(\theta)+1}$ .

You can start on either the left side or the right side. Either way, your aim is to reach the same thing as the other side. It is often easier to simplify an expression than make it less simple, but with this problem, starting from either side appears as difficult as the other. Express the term on the left in terms of $\sin(\theta)$ and $\cos(\theta)$ and work from there:

$\begin{aligned}\frac{\sin(\theta)}{1+\cot(\theta)}&\equiv\frac{\sin(\theta)}{1+\frac{\cos(\theta)}{\sin(\theta)}}\\&\equiv\frac{\sin^2(\theta)}{\sin(\theta)+\cos(\theta)}\\&\equiv\frac{1-\cos^2(\theta)}{\sin(\theta)+\cos(\theta)}\\&\equiv\frac{\frac1{\cos(\theta)}-\frac{\cos^2(\theta)}{\cos(\theta)}}{\frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\cos(\theta)}}\\&\equiv\underline{\frac{\sec(\theta)-\cos(\theta)}{\tan(\theta)+1}}\end{aligned}$