Class 12 · Chapter 2 · Relations and Functions unit (8 of 80 marks)

Inverse Trigonometric Functions Class 12: notes and important questions

Q: What is in CBSE Class 12 Maths Chapter 2, Inverse Trigonometric Functions?

The topics are: Principal values, Domain and range (principal value branches), Compositions and simplification, Graphs of inverse trigonometric functions. This page has revision notes for each, and 47 board-style practice questions with step-marked solutions.

Q: How many marks is Inverse Trigonometric Functions worth?

It is in the Relations and Functions unit, which carries 8 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Q: Are the Inverse Trigonometric Functions important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Q: Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 47 board-style questions.

Practise this chapter Join freeFree account, no card: +5 practice questions on every chapter, with instant feedback.

47 questions
15 multiple choice, 4 assertion–reason, 10 very short answer, 14 short answer, 1 long answer, 3 case study
About 10 hours to master

Relations and Functions unit: 8 of 80 theory marks (Relations and Functions and Inverse Trigonometric Functions).

NCERT textbook: Chapter 2 on ncert.nic.in (PDF)

Revision notes

Inverse Trigonometric Functions — revision notes

1. Principal value branches

Function	Domain	Range (principal branch)
$\sin^{- 1} x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$\cos^{- 1} x$	$[- 1, 1]$	$[0, π]$
$\tan^{- 1} x$	$R$	$(- \frac{π}{2}, \frac{π}{2})$
${cosec}^{- 1} x$	$R - (- 1, 1)$	$[- \frac{π}{2}, \frac{π}{2}] - {0}$
$\sec^{- 1} x$	$R - (- 1, 1)$	$[0, π] - {\frac{π}{2}}$
$\cot^{- 1} x$	$R$	$(0, π)$

$\sin^{- 1} x$ means “the angle in the principal range whose sine is $x$ ”. It is not $\frac{1}{\sin x}$ .

2. Graphs

The graph of an inverse function is the mirror image of the (restricted) original graph in the line $y = x$ . $\sin^{- 1}$ and $\tan^{- 1}$ are increasing and their graphs are symmetric about the origin (read this from the graph; the identity $\sin^{- 1} (- x) = - \sin^{- 1} x$ belongs to the deleted properties section); $\cos^{- 1}$ is decreasing from $π$ to $0$ ; $\tan^{- 1}$ has horizontal asymptotes $y = \pm \frac{π}{2}$ .

3. Compositions

$\sin (\sin^{- 1} x) = x$ for $x \in [- 1, 1]$ (similarly for the others on their domains).
$\sin^{- 1} (\sin θ) = θ$ only for $θ \in [- \frac{π}{2}, \frac{π}{2}]$ . Otherwise first rewrite $\sin θ = \sin ϕ$ with $ϕ$ in the principal range (using $\sin (π - θ) = \sin θ$ and period $2 π$ ); the answer is $ϕ$ .
Same idea for $\cos^{- 1} (\cos θ)$ (use $\cos (2 π - θ) = \cos θ$ ) and $\tan^{- 1} (\tan θ)$ (period $π$ ).
For $\tan (\cos^{- 1} \frac{3}{5})$ -type questions, draw a right triangle, but check the sign using the principal range.

Worked example 1

$\sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ , $\cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ .

Worked example 2

$\tan^{- 1} (\tan \frac{5 π}{6})$ : $\tan \frac{5 π}{6} = \tan (\frac{5 π}{6} - π) = \tan (- \frac{π}{6})$ , so the value is $- \frac{π}{6}$ .

Worked example 3

Domain of $\sin^{- 1} (2 x + 1)$ : $- 1 \leq 2 x + 1 \leq 1 \Rightarrow - 1 \leq x \leq 0$ .

Common errors

Writing $\sin^{- 1} (\sin \frac{2 π}{3}) = \frac{2 π}{3}$ — the answer must lie in the principal range.
Giving $\cos^{- 1} (- x)$ a negative value: $\cos^{- 1}$ is never negative.
Mixing up ranges of $\cot^{- 1}$ $(0, π)$ and $\tan^{- 1}$ $(- \frac{π}{2}, \frac{π}{2})$ .
Ignoring the sign of a square root after a substitution (e.g. $\sqrt{\cos^{2} θ} = | \cos θ |$ ).

Board-exam tips

State the principal range you are using in each step — it is where the marks are.
The general property formulas (sum of $\tan^{- 1}$ , $\sin^{- 1} x + \cos^{- 1} x$ , etc.) are no longer in the syllabus; reason from the branches instead.
Values in radians: $\sin 2$ means $2$ radians.

Topics in this chapter: Principal values · Domain and range (principal value branches) · Compositions and simplification · Graphs of inverse trigonometric functions.

Formulas for this chapter: Chapter 2 on the Class 12 formula sheet (free PDF and A3 poster).

Inverse Trigonometric Functions Class 12: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 5 pages · 227 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choicePrincipal values

What is the principal value of $\sec^{- 1} (- \sqrt{2})$ ?

Show worked answer

The principal branch of $\sec^{- 1}$ is $[0, π] - {\frac{π}{2}}$ , and $\sec \frac{3 π}{4} = - \sqrt{2}$ A1
Correct option: (b) $\frac{3 π}{4}$

Q2·1 mark·Multiple choiceDomain and range

The domain of $\sin^{- 1} (2 x - 1)$ is

Q3·1 mark·Assertion–reasonPrincipal values

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): $\sin^{- 1} (\sin \frac{3 π}{4}) = \frac{3 π}{4}$ .

Reason (R): The principal value branch of $\sin^{- 1}$ has range $[- \frac{π}{2}, \frac{π}{2}]$ .

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerEvaluating expressions

Find the value of $\cot^{- 1} (\sqrt{3}) + \sec^{- 1} (2) - \tan^{- 1} (- 1)$ .

Show worked answer

Answer:

\frac{3 π}{4}

$\cot^{- 1} \sqrt{3} = \frac{π}{6}$ , $\sec^{- 1} 2 = \frac{π}{3}$ , $\tan^{- 1} (- 1) = - \frac{π}{4}$ M1
Sum $= \frac{π}{6} + \frac{π}{3} + \frac{π}{4} = \frac{3 π}{4}$ A1

Q5·2 marks·Short answerPrincipal values

Find the principal value of $\cos^{- 1} (\cos \frac{5 π}{3})$ .

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerDomain and range

Find the domain of $f (x) = \sin^{- 1} (x^{2} - 3)$ .

Show worked answer

Answer:

[- 2, - \sqrt{2}] \cup [\sqrt{2}, 2]

Need $- 1 \leq x^{2} - 3 \leq 1$ , i.e. $2 \leq x^{2} \leq 4$ M1
$x^{2} \leq 4 \Rightarrow - 2 \leq x \leq 2$ ; $x^{2} \geq 2 \Rightarrow x \leq - \sqrt{2}$ or $x \geq \sqrt{2}$ A1
Domain $= [- 2, - \sqrt{2}] \cup [\sqrt{2}, 2]$ A1

Q7·3 marks·Short answerEvaluating expressions

Find the values of (a) $\cos (\tan^{- 1} \sqrt{3} + \cot^{- 1} \sqrt{3})$ and (b) $\tan (\sin^{- 1} \frac{\sqrt{3}}{2} - \cos^{- 1} \frac{\sqrt{3}}{2})$ .

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerPrincipal values

(a) Find the principal values of $\sin^{- 1} \frac{\sqrt{3}}{2}$ , $\cos^{- 1} (- \frac{1}{\sqrt{2}})$ , $\tan^{- 1} (- \sqrt{3})$ and ${cosec}^{- 1} (- 2)$ . (b) Hence find $\sin^{- 1} \frac{\sqrt{3}}{2} + \cos^{- 1} (- \frac{1}{\sqrt{2}}) - \tan^{- 1} (- \sqrt{3}) - {cosec}^{- 1} (- 2)$ .

Show worked answer

Answer:

\frac{π}{3}, \frac{3 π}{4}, - \frac{π}{3}, - \frac{π}{6}

;

\frac{19 π}{12}

$\sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3}$ (in $[- \frac{π}{2}, \frac{π}{2}]$ ) A1
$\cos^{- 1} (- \frac{1}{\sqrt{2}}) = \frac{3 π}{4}$ (in $[0, π]$ ) A1
$\tan^{- 1} (- \sqrt{3}) = - \frac{π}{3}$ (in $(- \frac{π}{2}, \frac{π}{2})$ ) A1
${cosec}^{- 1} (- 2) = - \frac{π}{6}$ (in $[- \frac{π}{2}, \frac{π}{2}] - {0}$ ) A1
(b) $\begin{aligned} \frac{π}{3} + \frac{3 π}{4} + \frac{π}{3} + \frac{π}{6} & = \frac{4 π + 9 π + 4 π + 2 π}{12} \\ = \frac{19 π}{12} \end{aligned}$ A1

Q9·5 marks·Long answerDomain and range

Let $f (x) = 2 \cos^{- 1} (1 - x)$ . (a) Find the domain and the range of $f$ . (b) Find $f (1)$ . (c) Solve $f (x) = \frac{2 π}{3}$ .

Section E style · Case study (4 marks)

Q10·4 marks·Case studyEvaluating expressions

Road gradients. A road rises $1$ m for every $1$ m travelled horizontally on one stretch, and $\sqrt{3}$ m for every $1$ m on a steeper stretch. The angle of a stretch with the horizontal is $θ = \tan^{- 1} (rise per metre)$ .

Find the angle of the first stretch. [1 mark]
State the range of the principal value branch of $\tan^{- 1}$ . [1 mark]
By how much is the steeper stretch steeper, in radians? [2 marks]

Show worked answer

$\tan^{- 1} 1 = \frac{π}{4}$ A1
$(- \frac{π}{2}, \frac{π}{2})$ A1
Steeper stretch: $\tan^{- 1} \sqrt{3} = \frac{π}{3}$ M1
Difference $= \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$ A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 12 assertion–reason questions · the Class 12 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can write the principal value branch of every inverse trig function and find principal values of standard values.

Read first: 1. Principal value branches; 2. Graphs 15 practice questions · checkpoint: 4 questions, 6 marks, pass 80%

Practise step 1

Step 2

Board standard

You can find domains, simplify compositions like sin⁻¹(sin x) using the principal range and evaluate expressions such as cos(tan⁻¹ ¾).

Read first: 1. Principal value branches; 3. Compositions; Worked examples 1-3 14 practice questions · checkpoint: 3 questions, 8 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can write complete multi-part answers (the case studies, the 5-mark simplification and the multi-part domain questions), stating the branch you use at each step.

Read first: 3. Compositions; Board-exam tips 6 practice questions · checkpoint: 3 questions, 13 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can handle radian inputs like sin⁻¹(sin 3), substitution-based simplifications and justifying the signs when you write sin θ and cos θ in terms of x.

Read first: 2. Graphs; 3. Compositions; Common errors 4 practice questions · checkpoint: 3 questions, 7 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 47 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choicePrincipal values

The principal value of $\sec^{- 1} (- \frac{2}{\sqrt{3}})$ is

(a) $- \frac{π}{6}$
(b) $\frac{5 π}{6}$
(c) $\frac{7 π}{6}$
(d) $\frac{2 π}{3}$

Q2·2 marks·Very short answerPrincipal values

If $\sin^{- 1} (2 x - 1) = \frac{π}{6}$ , find $x$ and hence the value of $\cos^{- 1} (1 - 2 x)$ .

Q3·1 mark·Multiple choiceDomain and range (principal value branches)

The range of the principal value branch of $\sec^{- 1} x$ is

(a) $[- \frac{π}{2}, \frac{π}{2}]$
(b) $[0, π] - {\frac{π}{2}}$
(c) $(0, π)$
(d) $[- \frac{π}{2}, \frac{π}{2}] - {0}$

Worked solution·2 marks·Very short answerPrincipal values

Find the value of $3 {cosec}^{- 1} (- 2) + 2 \cot^{- 1} (- \frac{1}{\sqrt{3}})$ .

Full worked solution Free, no sign-in

${cosec}^{- 1} (- 2) = - \frac{π}{6}$ (range $[- \frac{π}{2}, \frac{π}{2}] - {0}$ ); $\cot^{- 1} (- \frac{1}{\sqrt{3}}) = \frac{2 π}{3}$ (range $(0, π)$ ) M1
$3 (- \frac{π}{6}) + 2 \cdot \frac{2 π}{3} = - \frac{π}{2} + \frac{4 π}{3} = \frac{5 π}{6}$ A1

Every question here has a full worked solution. See them all →

Stretch yourself: original geometry problems at Senior level (ages 16 to 18), with hints and full solutions: Problem S74 · Problem S17 · Problem S73 · Problem S16. All 22 →

Knowledge organiser: Inverse Trigonometric Functions on one page (key facts, formulas and mistakes to avoid; printable PDF)

Where marks are lost in Inverse Trigonometric Functions

Writing sin⁻¹(sin 2π/3) = 2π/3. Fix: rewrite the angle into the principal range first (sin 2π/3 = sin π/3) and state the range.
Negative values for cos⁻¹ or cot⁻¹. Fix: cos⁻¹ lies in [0, π] and cot⁻¹ in (0, π); if you get a negative answer, re-check.
Mixing the ranges of tan⁻¹ and cot⁻¹. Fix: write the branch table at the top of your answer for any mixed question.
Dropping the modulus after a substitution (√cos²θ = |cos θ|). Fix: state the interval of θ and remove the modulus with the correct sign.
Using the deleted property formulas (tan⁻¹x + tan⁻¹y, etc.). Fix: they are out of the syllabus; reason from the principal branches and right-triangle values instead.
Treating sin 2 as sin 2°. Fix: numbers without a degree sign are radians; use 3 < π < 3.2, so π/2 is just over 1.5 and π is just over 3 (no calculator needed).

Exam strategy: how answers are step-marked · competency-based questions

Inverse Trigonometric Functions in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 2, 21
Sample paper 2 (Board standard): questions 2, 21
Sample paper 3 (Above board): questions 2, 21
Sample paper 4 (Above board): questions 2, 21
Sample paper 5 (95+ challenge): questions 2, 21

Want it against the clock? Take a timed 30-mark chapter test on Inverse Trigonometric Functions, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Preparing for JEE or CUET? Try the JEE and CUET-style questions on trigonometry and inverse trigonometric functions.

Inverse Trigonometric Functions: common questions

What is in CBSE Class 12 Maths Chapter 2, Inverse Trigonometric Functions?

The topics are: Principal values, Domain and range (principal value branches), Compositions and simplification, Graphs of inverse trigonometric functions. This page has revision notes for each, and 47 board-style practice questions with step-marked solutions.

How many marks is Inverse Trigonometric Functions worth?

It is in the Relations and Functions unit, which carries 8 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Inverse Trigonometric Functions?

Writing sin⁻¹(sin 2π/3) = 2π/3. Fix: rewrite the angle into the principal range first (sin 2π/3 = sin π/3) and state the range. Negative values for cos⁻¹ or cot⁻¹. Fix: cos⁻¹ lies in [0, π] and cot⁻¹ in (0, π); if you get a negative answer, re-check.

Are the Inverse Trigonometric Functions important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

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Revision notes

Inverse Trigonometric Functions — revision notes

1. Principal value branches

2. Graphs

3. Compositions

Worked example 1

Worked example 2

Worked example 3

Common errors

Board-exam tips

Inverse Trigonometric Functions Class 12: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Inverse Trigonometric Functions

Inverse Trigonometric Functions in our sample papers

Inverse Trigonometric Functions: common questions

Work through the Route to 95 with your progress saved