Trig Identities (AA HL)
HL trigonometry adds three new ratios, the reciprocals of sine, cosine and tangent, plus two extra Pythagorean identities built from them. Most exam marks here come from spotting which identity turns a messy expression into something that cancels. This page covers the definitions and identities with worked proofs and the mistakes that lose marks. It's part of the broader Identities & Equations topic.
11 questions on this sub-topic.
Reciprocal ratios and Pythagorean 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 \(1+\tan^2\theta=\sec^2\theta\) and \(1+\cot^2\theta=\csc^2\theta\). Both extra identities are listed in the formula booklet, but you still need to know how to derive them on the spot when a proof question asks for it.
Reciprocal ratios
\(\sec\theta = \dfrac{1}{\cos\theta}, \quad \csc\theta = \dfrac{1}{\sin\theta}, \quad \cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}\)
Each is undefined wherever its denominator ratio is zero - \(\sec\theta\) is undefined when \(\cos\theta=0\), and so on.
Extra Pythagorean identities
\(1+\tan^2\theta=\sec^2\theta \qquad 1+\cot^2\theta=\csc^2\theta\)
Both come from dividing \(\sin^2\theta+\cos^2\theta=1\) through by \(\cos^2\theta\) or \(\sin^2\theta\) - useful whenever a question mixes \(\tan\), \(\sec\), \(\cot\) or \(\csc\) in one expression.
Need the full syllabus wording and formula-booklet reference table? See Identities & Equations.
Worked examples
Given \(\sin\theta=\tfrac35\) with \(\theta\) acute, find \(\cos\theta\) and \(\tan\theta.\)
(a)(i) Find \(\cos\theta\).
(a)(ii) Find \(\tan\theta\).
Worked solution
(a)(i) \(= \tfrac45.\) A1
(a)(ii) \(\tan\theta = \dfrac{3/5}{4/5} = \tfrac34.\) A1
Prove \(\sec^2\theta+\csc^2\theta=\sec^2\theta\,\csc^2\theta.\)
Worked solution
\(\sec^2\theta + \csc^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta}.\) M1 A1
\(\dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta}.\) M1 Using \(\sin^2 + \cos^2 = 1\): \(= \dfrac{1}{\sin^2\theta\cos^2\theta}\) A1 \(= \sec^2\theta\csc^2\theta.\) A1 AG ∎
Prove \(\dfrac{1}{1-\sin\theta}+\dfrac{1}{1+\sin\theta}=\dfrac{2}{\cos^2\theta}.\)
Worked solution
\(\dfrac{(1+\sin\theta) + (1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\) M1 \(= \dfrac{2}{1 - \sin^2\theta}.\) A1
\(1 - \sin^2\theta = \cos^2\theta\) M1 so the expression \(= \dfrac{2}{\cos^2\theta}.\) A1 AG ∎
Simplify \(\dfrac{1}{\cos^2\theta}-\tan^2\theta.\)
Worked solution
\(\dfrac{1}{\cos^2\theta} = \sec^2\theta\), so the expression is \(\sec^2\theta - \tan^2\theta.\) M1 A1
\(\sec^2\theta - \tan^2\theta = 1.\) M1 A1
Common mistakes
- Mixing up which ratio is reciprocal to which. \(\sec\theta\) is \(1/\cos\theta\), not \(1/\sin\theta\) - the "co-" prefix on cosecant and cotangent is what pairs them with sine and tangent, which trips students who guess from the name.
- Forgetting the second Pythagorean identity exists. Many students only remember \(\sin^2\theta+\cos^2\theta=1\) and stall on a proof that actually needs \(1+\tan^2\theta=\sec^2\theta\) or \(1+\cot^2\theta=\csc^2\theta\) instead.
- Leaving an identity half-proved. "Show that" and "prove" questions need every algebraic step shown until both sides genuinely match - jumping straight to the answer without the intermediate common-denominator or substitution step loses method marks even if the final line is correct.
Ready to practise properly?
11 trig-identity 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 \(\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}\).
What are the extra Pythagorean identities at HL?
\(1+\tan^2\theta=\sec^2\theta\) and \(1+\cot^2\theta=\csc^2\theta\), both found by dividing \(\sin^2\theta+\cos^2\theta=1\) by \(\cos^2\theta\) or \(\sin^2\theta\) respectively. For guidance on entering these on your calculator, see the parent topic's GDC guidance.