Reciprocal Trig and Inverse Trig Functions (AA HL)
Alongside \(\sin\), \(\cos\) and \(\tan\), the HL syllabus adds their reciprocals - \(\sec\theta\), \(\csc\theta\) and \(\cot\theta\) - plus their own Pythagorean identities. This page covers what those ratios mean, how they connect to the identity you already know, and how to combine them with the inverse trig functions to solve equations. It's part of the broader Identities & Equations topic.
12 questions on this sub-topic.
Definitions and identities
Covered under IB syllabus reference AHL3.9: definition of the reciprocal trigonometric ratios \(\sec\theta\), \(\csc\theta\) and \(\cot\theta\), and the Pythagorean identities that come from them. The reciprocal ratio definitions aren't given in the booklet, so those need to be memorised - but the two Pythagorean identities are listed in the formula booklet.
Reciprocal ratios
\(\sec\theta=\dfrac{1}{\cos\theta}\), \(\csc\theta=\dfrac{1}{\sin\theta}\), \(\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}\)
Each is undefined exactly where its denominator ratio is zero - e.g. \(\sec\theta\) is undefined at \(\theta=\tfrac\pi2\), where \(\cos\theta=0\).
Pythagorean identities
\(1+\tan^2\theta=\sec^2\theta\)
\(1+\cot^2\theta=\csc^2\theta\)
Both come from dividing \(\sin^2\theta+\cos^2\theta=1\) by \(\cos^2\theta\) or \(\sin^2\theta\) respectively - useful for rewriting an equation with a mix of \(\tan\) and \(\sec\), or \(\cot\) and \(\csc\).
Need the full syllabus wording and formula-booklet reference table? See Identities & Equations.
Worked examples
\(1 + \tan^2\theta = \sec^2\theta.\)
(a) Prove that this holds.
(b) Find the exact value of \(\arctan(1)\), giving your answer in radians.
(c)(i) Solve \(\sec\theta = 2\) for \(0 \leq \theta \leq 2\pi\), giving the solution with \(\theta<3.142\).
(c)(ii) Give the solution with \(\theta>3.142\).
Worked solution
(a) Starting from \(\cos^2\theta + \sin^2\theta = 1\), divide through by \(\cos^2\theta\): M1
\(1 + \tan^2\theta = \sec^2\theta\) A1 AG
(b) \(\arctan(1) = \dfrac{\pi}{4}\) A1
(c)(i) \(\sec\theta = 2 \Rightarrow \cos\theta = \dfrac{1}{2}\) M1
\(\theta = \dfrac{\pi}{3}\) A1
(c)(ii) By symmetry, the other solution in \([0,2\pi]\) is \(\theta = \dfrac{5\pi}{3}\) A1
\(\theta\) lies in the third quadrant with \(\tan\theta=\tfrac{3}{4}\).
(a) Find \(\sin\theta\).
(b) Find \(\cos\theta\).
Worked solution
(a) \(\tan\theta=\tfrac34\) gives a 3-4-5 right triangle, so \(|\sin|=\tfrac35,\ |\cos|=\tfrac45\). M1
\(\sin\theta=-\tfrac35.\) A1
(b) In Q3 both sine and cosine are negative. R1 \(\cos\theta=-\tfrac45.\) A1
\(\cos\tfrac{\pi}{4}.\)
(a) Find its exact value.
(b) Find the exact value of \(\tan\tfrac{\pi}{3}\).
Worked solution
(a) Standard angle \(\tfrac{\pi}{4}=45^\circ\): \(\cos\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}.\) A1
(b) \(\tfrac{\pi}{3}=60^\circ\): \(\tan\tfrac{\pi}{3}\) A1
\(=\sqrt3.\) A1
Common mistakes
- Mixing up \(\sec\) and \(\csc\). \(\sec\theta\) is the reciprocal of \(\cos\theta\), not \(\sin\theta\) - the "s" in \(\sec\) doesn't line up with "sine" the way it looks like it should.
- Forgetting the sign from the quadrant. Knowing \(\tan\theta\) only gives you the magnitude of \(\sin\theta\) and \(\cos\theta\) - you still need to use which quadrant \(\theta\) is in (ASTC) to fix the correct signs.
- Applying the wrong Pythagorean identity. \(1+\tan^2\theta=\sec^2\theta\) pairs \(\tan\) with \(\sec\); \(1+\cot^2\theta=\csc^2\theta\) pairs \(\cot\) with \(\csc\) - swapping the pairing gives a false identity.
Ready to practise properly?
12 reciprocal/inverse-trig questions, marked instantly like the real exam.
Quick answers
What are the reciprocal trig ratios?
\(\sec\theta=\dfrac{1}{\cos\theta}\), \(\csc\theta=\dfrac{1}{\sin\theta}\), and \(\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}\). Each is undefined wherever its denominator ratio is zero.
What are the Pythagorean identities for sec and csc?
\(1+\tan^2\theta=\sec^2\theta\), and \(1+\cot^2\theta=\csc^2\theta\). Both follow from dividing \(\sin^2\theta+\cos^2\theta=1\) by \(\cos^2\theta\) or \(\sin^2\theta\) respectively. For calculator steps with inverse trig, see the parent topic's GDC guidance.