5.6 Secant, cosecant, cotangent and the inverse trigonometric functions — Practise

Work through these on your own. Give every value in exact form. For an inverse function, state the principal value, and give the angle in radians unless the question asks for degrees.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Write down the value of $\sec\dfrac{\pi}{3}$ .
Write down the value of $\operatorname{cosec}\dfrac{\pi}{6}$ .
Write down the value of $\cot\dfrac{\pi}{4}$ .
Simplify $\operatorname{cosec}^2\theta - \cot^2\theta$ .
Simplify $\sec^2\theta - \tan^2\theta$ .
Given that $\cos\theta = \dfrac{5}{13}$ and $\theta$ is acute, find $\sin\theta$ and $\operatorname{cosec}\theta$ .
Work out $\arcsin\dfrac{\sqrt{3}}{2}$ , giving the principal value in radians.
Work out $\arccos\dfrac{1}{2}$ , giving the principal value in radians.
Work out $\arctan 1$ , giving the principal value in radians.
Given that $\sec\theta = \dfrac{5}{3}$ and $\theta$ is acute, find $\tan\theta$ .

Answers and concise solutions

1. The value

Answer: $\sec\dfrac{\pi}{3} = \dfrac{1}{\cos\dfrac{\pi}{3}} = \dfrac{1}{\frac{1}{2}} = 2$ .

2. The value

Answer: $\operatorname{cosec}\dfrac{\pi}{6} = \dfrac{1}{\sin\dfrac{\pi}{6}} = \dfrac{1}{\frac{1}{2}} = 2$ .

3. The value

Answer: $\cot\dfrac{\pi}{4} = \dfrac{\cos\dfrac{\pi}{4}}{\sin\dfrac{\pi}{4}} = \dfrac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$ .

4. The simplified expression

Answer: From $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ , subtract $\cot^2\theta$ from both sides. The result is $\operatorname{cosec}^2\theta - \cot^2\theta = 1$ .

5. The simplified expression

Answer: From $\sec^2\theta = 1 + \tan^2\theta$ , subtract $\tan^2\theta$ from both sides. The result is $\sec^2\theta - \tan^2\theta = 1$ .

6. The two values

Answer: $\sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$ . Since $\theta$ is acute, $\sin\theta = \dfrac{12}{13}$ . Then $\operatorname{cosec}\theta = \dfrac{13}{12}$ .

7. The principal value

Answer: The principal range of arcsine is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . Inside it, $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ , so $\arcsin\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{3}$ .

8. The principal value

Answer: The principal range of arccosine is $[0, \pi]$ . Inside it, $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$ , so $\arccos\dfrac{1}{2} = \dfrac{\pi}{3}$ .

9. The principal value

Answer: The principal range of arctangent is $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ . Inside it, $\tan\dfrac{\pi}{4} = 1$ , so $\arctan 1 = \dfrac{\pi}{4}$ .

10. The value

Answer: Use $\sec^2\theta = 1 + \tan^2\theta$ . Then $\tan^2\theta = \dfrac{25}{9} - 1 = \dfrac{16}{9}$ , so $\tan\theta = \dfrac{4}{3}$ because $\theta$ is acute.