Revision notes
Reciprocal and Pythagorean identities in trigonometry
cot θ, sec θ, cosec θ
Hint: look at the third letter of each term
$ \cot \theta = $ $ \phantom{.} {1 \over \tan \theta} $,
$ \sec \theta = $ $ \phantom{.} {1 \over \cos \theta} $,
$ \text{cosec } \theta = $ $ \phantom{.} {1 \over \sin \theta} $
sin θ, cos θ and tan θ
$ \tan \theta = $ $ \phantom{.} {\sin \theta \over \cos \theta} $,
$ \cot \theta = $ $ \phantom{.} {1 \over \tan \theta} = {\cos \theta \over \sin \theta} $
Pythagorean identities (provided)
\begin{align*}
\sin^2 A & + \cos^2 A = 1 \phantom{000} (1) \\
\sec^2 A & = 1 + \tan^2 A \phantom{000} (2) \\
\text{cosec}^2 A & = 1 + \cot^2 A \phantom{000} (3) \\
\end{align*}
Manipulations of identity:
From $(1)$, $\sin^2 A =$ $ 1 - \cos^2 A $
From $(1)$, $\cos^2 A =$ $ 1 - \sin^2 A $
From $(2)$, $\tan^2 A =$ $ \sec^2 A - 1 $
From $(3)$, $\cot^2 A =$ $ \text{cosec}^2 A - 1 $
Practice questions
Prove a trigonometric identity
Prove identity by factorisation
Identity used for factorisation:
$ a^2 - b^2 = $$ (a + b)(a - b) $
For example, $1 - \sin^2 \theta$ = $ 1^2 - (\sin \theta)^2 = (1 + \sin \theta)(1 - \sin \theta) $
1. Prove the following identities
(i) $ \sin^2 4 \theta - \cos^4 \theta = 2 \sin^2 \theta - 1 $
Solutions
\begin{align*}
\text{L.H.S} & = \sin^4 \theta - \cos^4 \theta \\
& = (\sin^2 \theta)^2 - (\cos^2 \theta)^2 \\
& = (\sin^2 \theta + \cos^2 \theta)(\sin^2 \theta - \cos^2 \theta) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\
& = (1)(\sin^2 \theta - \cos^2 \theta) \phantom{00000000000000000} [ \sin^2 A + \cos^2 A = 1] \\
& = \sin^2 \theta - \cos^2 \theta \\
& = \sin^2 \theta - (1 - \sin^2 \theta) \phantom{0000000000000000.} [\cos^2 A = 1- \sin^2 A] \\
& = \sin^2 \theta - 1 + \sin^2 \theta \\
& = 2 \sin^2 \theta - 1 \\
& = \text{R.H.S}
\end{align*}
(ii) $ \tan^2 x - \sin^2 x = \tan^2 x \sin^2 x $
Solutions
\begin{align*}
\text{L.H.S} & = \tan^2 x - \sin^2 x \\
& = {\sin^2 x \over \cos^2 x} - {\sin^2 x \over 1} \\
& = {\sin^2 x \over \cos^2 x} - {\sin^2 x \cos^2 x \over \cos^2 x} \\
& = {\sin^2 x - \sin^2 x \cos^2 x \over \cos^2 x} \\
& = {\sin^2 x (1 - \cos^2 x) \over \cos^2 x} \\
& = {\sin^2 x (\sin^2 x) \over \cos^2 x} \phantom{0000000000000000.} [\sin^2 A = 1- \cos^2 A] \\
& = \left(\sin^2 x \over \cos^2 x\right) \left(\sin^2 x \over 1\right) \\
& = \tan^2 x \sin^2 x \\
& = \text{R.H.S}
\end{align*}
Prove identity involving sin, cos and tan
2. Prove the identity $ {\tan \theta + 1 \over \tan \theta - 1} = {\sin \theta + \cos \theta \over \sin \theta - \cos \theta} $.
Solutions
\begin{align*}
\text{L.H.S} & = {\tan \theta + 1 \over \tan \theta - 1} \\
& = { {\sin \theta \over \cos \theta} + 1 \over {\sin \theta \over \cos \theta} - 1} \times {\cos \theta \over \cos \theta} \\
& = { \cos \theta \left( {\sin \theta \over \cos \theta} + 1 \right) \over \cos \theta \left( {\sin \theta \over \cos \theta} - 1 \right) } \\
& = { \sin \theta + \cos \theta \over \sin \theta - \cos \theta } \\
& = \text{R.H.S}
\end{align*}
Prove identity by combining fractions
3. Prove the following identities
(i) $ {\sin x \over 1 - \cos x} + {1 - \cos x \over \sin x} = 2 \text{cosec } x $
Solutions
\begin{align*}
\require{cancel}
\text{L.H.S} & = {\sin x \over 1 - \cos x} + {1 - \cos x \over \sin x} \\
& = {\sin x(\sin x) \over \sin x(1 - \cos x) } + {(1 - \cos x)(1 - \cos x) \over \sin x(1 - \cos x)} \\
& = {\sin^2 x + (1 - \cos x)^2 \over \sin x(1 - \cos x) } \\
& = {\sin^2 x + 1 - 2(1)(\cos x) + (\cos x)^2 \over \sin x(1 - \cos x)} \phantom{000000} [ (a - b)^2 = a^2 - 2ab + b^2 ] \\
& = {\sin^2 x + 1 - 2 \cos x + \cos^2 x \over \sin x(1 - \cos x) } \\
& = {1 + 1 - 2 \cos x \over \sin x(1 - \cos x)} \phantom{0000000000000000000000} [\sin^2 A + \cos^2 A = 1] \\
& = {2 - 2 \cos x \over \sin x(1 - \cos x)} \\
& = {2 \cancel{(1 - \cos x)} \over \sin x \cancel{(1 - \cos x} )} \\
& = {2 \over \sin x} \\
& = \left(2 \over 1\right) \left(1 \over \sin x\right) \\
& = 2 \text{cosec } x \\
& = \text{R.H.S}
\end{align*}
(ii) $ {cos x \over 1 + \cos x} - {\cos x \over 1 - \cos x} = - 2\cot^2 x $
Solutions
\begin{align*}
\text{L.H.S} & = {\cos x \over 1 + \cos x} - {\cos x \over 1 - \cos x} \\
& = {\cos x (1 - \cos x) \over (1 + \cos x)(1 - \cos x)} - {\cos x (1 + \cos x) \over (1 + \cos x)(1 - \cos x)} \\
& = {\cos x(1 - \cos x) - \cos x(1 + \cos x) \over (1 + \cos x)(1 - \cos x)} \\
& = {\cos x - \cos^2 x - \cos x - \cos^2 x \over 1^2 - (\cos x)^2} \phantom{0000000} [(a + b)(a - b) = a^2 - b^2] \\
& = {- 2 \cos^2 x \over 1 - \cos^2 x} \\
& = {- 2 \cos^2 x \over \sin^2 x} \phantom{0000000000000000000000000} [\sin^2 A = 1 - \cos^2 A] \\
& = \left(-2 \over 1\right) \left(\cos^2 x \over \sin^2 x\right) \\
& = - 2 \cot^2 x \\
& = \text{R.H.S}
\end{align*}
Solve trigonometric equations using identities
4. Solve the equation $2 \tan^2 x + 3 \sec x = 0 $ for $0^\circ < x < 360^\circ$.
Answers: $ 120^\circ, 240^\circ $
Solutions
\begin{align*}
2 \tan^2 x + 3 \sec x & = 0 \\
2(\sec^2 x - 1) + 3 \sec x & = 0 \phantom{000000} [\tan^2 A = \sec^2 A - 1] \\
2 \sec^2 x - 2 + 3 \sec x & = 0 \\
2 \sec^2 x + 3 \sec x - 2 & = 0 \phantom{000000} [\text{Quadratic equation in } \sec x] \\
(2 \sec x - 1) (\sec x + 2) & = 0
\end{align*}
\begin{align*}
\sec x + 2 & = 0 && \text{ or } & 2 \sec x - 1 & = 0 \\
\sec x & = -2 &&& 2 \sec x & = 1 \\
{1 \over \cos x} & = -2 &&& \sec x & = {1 \over 2} \\
1 & = -2 \cos x &&& {1 \over \cos x} & = {1 \over 2} \\
-{1 \over 2} & = \cos x &&& \cos x & = 2 \phantom{0} \text{ (No solutions since } -1 \le \cos x \le 1)
\end{align*}
\begin{align*}
\cos x & = -{1 \over 2} \phantom{000000} [\text{2nd or 3rd quadrant since } \cos x < 0] \\
\\
\text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\
& = 60^\circ
\end{align*}
\begin{align*}
x & = 180^\circ - 60^\circ, 180^\circ + 60^\circ \\
& = 120^\circ, 240^\circ
\end{align*}
Prove identity then use it to solve an equation
5(i) Prove the identity $(\sec x + \tan x)^2 = {1 + \sin x \over 1 - \sin x} $
Solutions
\begin{align*}
\require{cancel}
\text{L.H.S} & = (\sec x + \tan x)^2 \\
& = \left( {1 \over \cos x} + {\sin x \over \cos x} \right)^2 \\
& = \left( {1 + \sin x \over \cos x} \right)^2 \\
& = { (1 + \sin x)^2 \over (\cos x)^2 } \\
& = { (1 + \sin x)^2 \over \cos^2 x} \\
& = { (1 + \sin x)^2 \over 1 - \sin^2 x} \phantom{000000000000000} [\cos^2 A = 1 - \sin^2 A] \\
& = { (1 + \sin x)^\cancel{2} \over \cancel{(1 + \sin x)}(1 - \sin x)} \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\
& = { 1 + \sin x \over 1 - \sin x} \\
& = \text{R.H.S}
\end{align*}
5(ii) Hence, solve the equation $ (\sec 2 \theta + \tan 2\theta)^2 = 5 $ for $0 < \theta < 6$.
Answers: $ 0.365, 1.21, 3.51, 4.35 $
Solutions
\begin{align*}
\text{From (i), } (\sec x + \tan x)^2 & = {1 + \sin x \over 1 - \sin x} \\
\implies (\sec 2 \theta + \tan 2 \theta)^2 & = {1 + \sin 2 \theta \over 1 - \sin 2 \theta} \\
\\
\therefore {1 + \sin 2 \theta \over 1 - \sin 2 \theta} & = 5 \\
1 + \sin 2 \theta & = 5(1 - \sin 2 \theta) \\
1 + \sin 2 \theta & = 5 - 5 \sin 2 \theta \\
6 \sin 2 \theta & = 4 \\
\sin 2 \theta & = {4 \over 6} = {2 \over 3} \phantom{000000} [\text{1st or 2nd quadrant}] \\
\\
\text{Basic angle, } \alpha & = \sin^{-1} \left(2 \over 3\right) \\
& = 0.72972 \text{ radians}
\end{align*}
\begin{align*}
[& \text{Since } 0 < \theta < 6, \phantom{0} 0 < 2 \theta < 12 ] \\
\\
2 \theta & = 0.72972, \pi - 0.72972 \\
& = 0.72972, 2.4118, 0.72972 + 2\pi, 2.4118 + 2\pi \\
& = 0.72972, 2.4118, 7.0129, 8.6949 \\
\\
\theta & = 0.36486, 1.2059, 3.5064, 4.3474 \\
& \approx 0.365, 1.21, 3.51, 4.35
\end{align*}
O Level past year questions on trigonometric identities
Fully worked, step-by-step solutions to these past-year questions (2016 to 2025) are in the O Level A Maths Solutions page. For 2015 and earlier, selected questions and their solutions are available to subscribers, linked individually in the table below.
| Year & paper |
Comments |
| 2025 P2 Question 6 |
Solve equation |
| 2023 P1 Question 3 |
Prove identity |
| 2023 P1 Question 5 |
Solve equation |
| 2021 P1 Question 10 |
Prove identity in first part, then use identity to solve equation in second part |
| 4049 Specimen P1 Question 12 |
Prove identity in first part, then use identity to solve equation in second part |
| 2020 P1 Question 7 |
Solve equation |
| 2019 P2 Question 2 |
Prove identity in first part, then use identity to solve equation in second part |
| 2017 P1 Question 5 |
Prove identity in first part, then use identity to solve equation in second part |
| 2016 P1 Question 11a |
Prove identity |
| 2015 P1 Question 8 |
Solve equation |
| 2014 P1 Question 6 |
Prove identity in first part, then use identity to solve equation in second part |
| 2013 P1 Question 4 |
Prove identity in first part, then use identity to solve equation in second part (Link 🔒 Subscribers) |
| 2010 P2 Question 1 |
Solve equation |
| 2007 P2 Question 1 |
Prove identity |
| 2007 P2 Question 9 |
Solve equation |
| 2006 P1 Question 11ii |
Solve equation |
| 2006 P2 Question 2 |
Prove identity |
| 2005 P1 Question 9a |
Solve equation (Link 🔒 Subscribers) |
| 2004 P1 Question 9a |
Solve equation |
| 2004 P2 Question 6 |
Prove identity (Link 🔒 Subscribers) |
| 2003 P1 Question 9i |
Solve equation |
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