13.12 Reciprocal Trigonometric Identities
AHL 3.9
This section is for HL students. It builds on 13.11.
- the reciprocal identities for \(\sec\theta\) and \(\operatorname{cosec}\theta\), and the identity \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\)
- the identities \(1 + \tan^2\theta = \sec^2\theta\) and \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\)
- how to use them to find ratios, simplify expressions and prove identities
Reciprocal identities
The three reciprocal ratios were introduced in Section 13.9.
Reciprocal trigonometric identities In the HL formula booklet
\[ \sec\theta = \frac{1}{\cos\theta}, \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \]
- \(\sec\theta\): the secant of \(\theta\) (defined when \(\cos\theta \neq 0\))
- \(\operatorname{cosec}\theta\): the cosecant of \(\theta\) (defined when \(\sin\theta \neq 0\))
Identity for \(\cot\theta\) Not in the formula booklet
\[ \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}, \qquad \sin\theta \neq 0 \]
- \(\cot\theta\): the cotangent of \(\theta\)
The second form, \(\dfrac{\cos\theta}{\sin\theta}\), also works when \(\cos\theta = 0\) (then \(\cot\theta = 0\), although \(\tan\theta\) is not defined).
Example 1 Given that \(\sin\theta = \dfrac{5}{13}\) and \(\cos\theta = -\dfrac{12}{13}\), so that \(\tan\theta = -\dfrac{5}{12}\) (13.11), find the value of
(a) \(\cot\theta\)
(b) \(\sec\theta\)
(a) \(\cot\theta = \dfrac{1}{\tan\theta} = -\dfrac{12}{5}\).
(b) \(\sec\theta = \dfrac{1}{\cos\theta} = -\dfrac{13}{12}\).
Pythagorean identities
Start from the Pythagorean identity of 13.11:
\[ \cos^2\theta + \sin^2\theta = 1 . \qquad (2) \]
Divide both sides of (2) by \(\cos^2\theta\):
\[ \frac{\cos^2\theta}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \;\Longrightarrow\; 1 + \tan^2\theta = \sec^2\theta . \]
Divide both sides of (2) by \(\sin^2\theta\):
\[ \frac{\cos^2\theta}{\sin^2\theta} + \frac{\sin^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} \;\Longrightarrow\; \cot^2\theta + 1 = \operatorname{cosec}^2\theta . \]
Pythagorean identities (AHL) In the HL formula booklet
\[ 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \operatorname{cosec}^2\theta \]
- \(\theta\): any angle where both sides are defined
- \(\tan^2\theta\), \(\sec^2\theta\), \(\cot^2\theta\), \(\operatorname{cosec}^2\theta\): the squares of \(\tan\theta\), \(\sec\theta\), \(\cot\theta\), \(\operatorname{cosec}\theta\)
You do not need to learn these two by heart: divide \(\cos^2\theta + \sin^2\theta = 1\) by \(\cos^2\theta\) or by \(\sin^2\theta\).
Simplifying expressions
To simplify, it often helps to write everything in terms of \(\sin\) and \(\cos\), and to look for \(\sec^2\theta - 1\) and \(\operatorname{cosec}^2\theta - 1\).
Example 2 Simplify each expression.
(a) \(\sin\theta + \cos\theta\cot\theta\)
(b) \(\sec^2\theta\tan\theta - \tan\theta\)
(c) \(\dfrac{\tan^2\theta}{1 - \cos^2\theta}\)
(a) Write \(\cot\theta\) as \(\dfrac{\cos\theta}{\sin\theta}\) and use a common denominator:
\[ \sin\theta + \frac{\cos^2\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta} = \frac{1}{\sin\theta} = \operatorname{cosec}\theta . \]
(b) Take out the common factor \(\tan\theta\):
\[ \tan\theta\left(\sec^2\theta - 1\right) = \tan\theta \times \tan^2\theta = \tan^3\theta . \]
(c) \(1 - \cos^2\theta = \sin^2\theta\), so
\[ \begin{aligned} \frac{\tan^2\theta}{\sin^2\theta} &= \frac{\sin^2\theta}{\cos^2\theta} \times \frac{1}{\sin^2\theta} \\ &= \frac{1}{\cos^2\theta} \\ &= \sec^2\theta . \end{aligned} \]
Proving identities
Use the method for proving identities from 13.11: start with one side and work towards the other.
Example 3 Prove each identity.
(a) \((1 - \cos^2 x)(1 + \tan^2 x) \equiv \tan^2 x\)
(b) \(\dfrac{1 - \sin\theta}{\sin\theta} \equiv \dfrac{\cot^2\theta}{\operatorname{cosec}\theta + 1}\)
(a) Use \(1 - \cos^2 x = \sin^2 x\) and \(1 + \tan^2 x = \sec^2 x\):
\[ \text{LHS} = \sin^2 x \times \sec^2 x = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x = \text{RHS} . \]
(b) Start with the RHS (it looks more complicated). Use \(\cot^2\theta = \operatorname{cosec}^2\theta - 1\) and factorise:
\[ \begin{aligned} \text{RHS} &= \frac{\operatorname{cosec}^2\theta - 1}{\operatorname{cosec}\theta + 1} = \frac{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}{\operatorname{cosec}\theta + 1} \\ &= \operatorname{cosec}\theta - 1 = \frac{1}{\sin\theta} - 1 = \frac{1 - \sin\theta}{\sin\theta} = \text{LHS} . \end{aligned} \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\tan^2\theta + 1 = \operatorname{cosec}^2\theta\) | \(\tan^2\theta + 1 = \sec^2\theta\) | Divide by \(\cos^2\theta\) to check. |
Exercises
1 Prove that \(\sec\theta - \cos\theta \equiv \sin\theta\tan\theta\).
\[ \begin{aligned} \text{LHS} &= \frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} \\ &= \sin\theta \times \frac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta = \text{RHS} . \end{aligned} \]
2 Given that \(\sec\theta = 3\) and \(\dfrac{3\pi}{2} < \theta < 2\pi\), find the exact values of \(\tan\theta\) and \(\operatorname{cosec}\theta\).
\(\tan^2\theta = \sec^2\theta - 1 = 8\). In the fourth quadrant \(\tan\theta < 0\), so \(\tan\theta = -2\sqrt{2}\).
\(\cos\theta = \dfrac{1}{3}\), so \(\sin\theta = \tan\theta\cos\theta = -\dfrac{2\sqrt{2}}{3}\) and
\[ \operatorname{cosec}\theta = -\frac{3}{2\sqrt{2}} = -\frac{3\sqrt{2}}{4} . \]