MathsCBSE Class 12 · Inverse Trigonometric Functions

Domain and Range of Inverse Trigonometric Functions (Class 12)

Updated 15 October 2026 · Fact-checked

sin⁻¹x and cos⁻¹x need −1 ≤ x ≤ 1, with ranges [−π/2, π/2] and [0, π]. tan⁻¹x and cot⁻¹x accept every real x, with ranges (−π/2, π/2) and (0, π). sec⁻¹x and cosec⁻¹x need |x| ≥ 1. For sin⁻¹(g(x)), solve −1 ≤ g(x) ≤ 1.

Understand Domain and Range of Inverse Trigonometric Functions

A trigonometric function such as $sin x$ is not one-one on $R$ : $sin 0 = sin π = 0$ . To define an inverse, we restrict it to an interval where it is one-one and still takes every value it can. That interval, the principal value branch, becomes the range of the inverse function, and the set of values the original function takes becomes its domain. So for each inverse function, the domain and range swap roles with the restricted trigonometric function.

The six standard results (NCERT Section 2.2):

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

Why these domains? $sin θ$ and $cos θ$ only take values between $- 1$ and $1$ , so $sin^{- 1} x$ and $cos^{- 1} x$ only accept such $x$ . $tan θ$ and $cot θ$ take every real value. $sec θ = \frac{1}{c o s θ}$ has $∣ sec θ ∣ \geq 1$ , so $sec^{- 1} x$ and $cosec^{- 1} x$ accept only $x \leq - 1$ or $x \geq 1$ .

Why these ranges? Each branch is the simplest interval containing the angles from 0 to $\frac{π}{2}$ on which the function is one-one and takes all its values. The 'sin family' ( $sin^{- 1}$ , $tan^{- 1}$ , $cosec^{- 1}$ ) is centred on 0, from $- \frac{π}{2}$ to $\frac{π}{2}$ . The 'cos family' ( $cos^{- 1}$ , $cot^{- 1}$ , $sec^{- 1}$ ) runs from 0 to $π$ . Ends are left out where the function would be undefined: $tan$ at $\pm \frac{π}{2}$ , $cot$ at $0$ and $π$ , $sec$ at $\frac{π}{2}$ , $cosec$ at $0$ .

Composite domains. For $sin^{- 1} (g (x))$ or $cos^{- 1} (g (x))$ , the inside must lie in $[- 1, 1]$ : solve $- 1 \leq g (x) \leq 1$ as two inequalities and intersect. For $sec^{- 1} (g (x))$ or $cosec^{- 1} (g (x))$ , solve $∣ g (x) ∣ \geq 1$ , which gives a union of two pieces. For a sum such as $sin^{- 1} x + sec^{- 1} x$ , intersect the separate domains.

Ranges of transformed functions. Because $sin^{- 1}$ and $tan^{- 1}$ are increasing and $cos^{- 1}$ and $cot^{- 1}$ are decreasing, you get the range of $a sin^{- 1} x + b$ by applying the same operations to the ends of the basic range (reversing the ends when $a < 0$ ). Over a restricted interval of $x$ , evaluate at the two ends.

These facts also give the graphs: the graph of $y = sin^{- 1} x$ is the mirror image of the principal branch of $y = sin x$ in the line $y = x$ .

Domain and Range of Inverse Trigonometric Functions: formulas to remember

Inverse sine: $sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$; Increasing; $sin^{- 1} (- x) = - sin^{- 1} x$ follows from the symmetric branch.
Inverse cosine: $cos^{- 1} : [- 1, 1] \to [0, π]$; Decreasing; never negative.
Inverse tangent: $tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2})$; Open interval: the ends are never reached.
Inverse cotangent: $cot^{- 1} : R \to (0, π)$; Decreasing; open interval.
Inverse secant: $sec^{- 1} : R - (- 1, 1) \to [0, π] - {\frac{π}{2}}$; Domain: $x \leq - 1$ or $x \geq 1$ .
Inverse cosecant: $cosec^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}] - {0}$; Domain: $∣ x ∣ \geq 1$ .
Composite domain rule: $sin^{- 1} (g (x)), cos^{- 1} (g (x)) defined ⟺ - 1 \leq g (x) \leq 1$; For $sec^{- 1}$ and $cosec^{- 1}$ : $∣ g (x) ∣ \geq 1$ .

How to solve Domain and Range of Inverse Trigonometric Functions questions

Use these steps for 'find the domain' and 'find the range' questions, including MCQs on graphs.

1Identify the outer inverse function and write its domain condition on the inside expression: $- 1 \leq g (x) \leq 1$ for $sin^{- 1}$ and $cos^{- 1}$ ; $∣ g (x) ∣ \geq 1$ for $sec^{- 1}$ and $cosec^{- 1}$ ; no condition for $tan^{- 1}$ and $cot^{- 1}$ .
2Solve the inequality. For a double inequality with $x^{2}$ , split it into two inequalities, solve each, and intersect the solution sets.
3If there are other parts (square roots, denominators, a second inverse function), write their conditions too and intersect everything.
4Write the domain in interval notation, with square brackets for included ends and round brackets for excluded ends.
5For a range, start from the basic range of the inverse function and apply the given operations (multiply, add) to both ends, reversing the order if you multiply by a negative number.
6If $x$ is restricted to an interval, evaluate the inverse function at both ends; since each inverse function is monotonic, the range lies between these values.

Domain and Range of Inverse Trigonometric Functions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Domain of the inverse sine

Suppose a school builds a wheelchair ramp from a straight board 4 m long. If the top of the ramp is $h$ m above the ground, the angle of the ramp with the ground is $θ = sin^{- 1} (\frac{h}{4})$ .

For what heights $h$ does this formula make sense mathematically? The school wants the angle to be at most $\frac{π}{6}$ . What is the greatest height the ramp may reach? What is the angle when $h = 22$ m?

Try it first, then show the solutionSolution

$sin^{- 1}$ needs $- 1 \leq \frac{h}{4} \leq 1$ , that is $- 4 \leq h \leq 4$ . For a real ramp, $0 \leq h \leq 4$ .
$sin^{- 1}$ is increasing, so $sin^{- 1} \frac{h}{4} \leq \frac{π}{6} ⟺ \frac{h}{4} \leq sin \frac{π}{6} = \frac{1}{2} ⟺ h \leq 2$ .
For $h = 22$ : $θ = sin^{- 1} \frac{2 2}{4} = sin^{- 1} \frac{1}{2} = \frac{π}{4}$ .

Answer: Mathematically $- 4 \leq h \leq 4$ (physically $0 \leq h \leq 4$ ). The greatest height is 2 m. At $h = 22$ m the angle is $\frac{π}{4}$ , which is too steep.

Memory hook: A ramp can never be higher than it is long: that is why $sin^{- 1}$ only accepts inputs up to 1.

Range of the inverse tangent

Imagine a camera on a tripod pointing at a bird perched 3 m above the camera lens. If the horizontal distance to the bird is $d$ m, the camera must tilt up by $θ = tan^{- 1} (\frac{3}{d})$ .

Find the tilt for $d = 33$ , $d = 3$ and $d = 3$ . Explain why the tilt is always less than $\frac{π}{2}$ , however close the bird is.

Try it first, then show the solutionSolution

$d = 33$ : $tan^{- 1} \frac{1}{3} = \frac{π}{6}$ .
$d = 3$ : $tan^{- 1} 1 = \frac{π}{4}$ .
$d = 3$ : $tan^{- 1} 3 = \frac{π}{3}$ .
The range of $tan^{- 1}$ is the open interval $(- \frac{π}{2}, \frac{π}{2})$ . As $d$ gets smaller, $\frac{3}{d}$ grows without bound and $θ$ gets closer to $\frac{π}{2}$ but never equals it.

Answer: $\frac{π}{6}$ , $\frac{π}{4}$ and $\frac{π}{3}$ ; the tilt stays below $\frac{π}{2}$ because $\frac{π}{2}$ is not in the range of $tan^{- 1}$ .

Memory hook: The camera never points straight up: the range of $tan^{- 1}$ is open at $\pm \frac{π}{2}$ .

Tricks to remember Domain and Range of Inverse Trigonometric Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sin family centred, cos family from zero
$sin^{- 1}$ , $tan^{- 1}$ , $cosec^{- 1}$ live around zero: $- \frac{π}{2}$ to $\frac{π}{2}$ . $cos^{- 1}$ , $cot^{- 1}$ , $sec^{- 1}$ live from 0 to $π$ .
Use it when: Writing the range column of the table from memory.
Brackets follow 'undefined'
Leave out an end only where the original function is undefined: $tan$ at $\pm \frac{π}{2}$ , $cot$ at 0 and $π$ , $sec$ at $\frac{π}{2}$ , $cosec$ at 0.
Use it when: Choosing between square and round brackets in MCQs.
Sine and cosine stay inside the unit band
Anything inside $sin^{- 1}$ or $cos^{- 1}$ must sit between $- 1$ and $1$ . Anything inside $sec^{- 1}$ or $cosec^{- 1}$ must sit outside that band.
Use it when: Setting up composite-domain inequalities.
Increasing pair, decreasing pair
$sin^{- 1}$ and $tan^{- 1}$ go up; $cos^{- 1}$ and $cot^{- 1}$ go down. For a range over an interval, plug in the ends and order them.
Use it when: Range questions over restricted intervals.

Domain and Range of Inverse Trigonometric Functions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A linear expression inside sin inverse · 2 marks · Very short answerBoard level

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

Try it first, then show the solutionSolution

$sin^{- 1}$ is defined only for inputs in $[- 1, 1]$ , so $- 1 \leq 2 x - 1 \leq 1$ . (1 mark)
Add 1: $0 \leq 2 x \leq 2$ ; divide by 2: $0 \leq x \leq 1$ . (1 mark)

Answer: Domain $= [0, 1]$ .

Example 2: Domain of an inverse secant · 2 marks · Very short answerBoard level

Find the domain of $f (x) = sec^{- 1} (2 x + 1)$ .

Try it first, then show the solutionSolution

$sec^{- 1}$ needs $∣2 x + 1∣ \geq 1$ . (½ mark)
So $2 x + 1 \leq - 1$ or $2 x + 1 \geq 1$ . (½ mark)
That is $x \leq - 1$ or $x \geq 0$ . (1 mark)

Answer: Domain $= (- \infty, - 1] \cup [0, \infty)$ .

Example 3: A quadratic inside cos inverse · 3 marks · Short answerHarder board level

Find the domain of $f (x) = cos^{- 1} (x^{2} - 4)$ .

Try it first, then show the solutionSolution

Need $- 1 \leq x^{2} - 4 \leq 1$ , that is $3 \leq x^{2} \leq 5$ . (1 mark)
$x^{2} \geq 3 \Rightarrow x \leq - 3$ or $x \geq 3$ . (½ mark)
$x^{2} \leq 5 \Rightarrow - 5 \leq x \leq 5$ . (½ mark)
Intersecting the two conditions: $- 5 \leq x \leq - 3$ or $3 \leq x \leq 5$ . (1 mark)

Answer: Domain $= [- 5, - 3] \cup [3, 5]$ .

Example 4: Range of a transformed inverse sine · 2 marks · Very short answerHarder board level

Find the range of $f (x) = 2 sin^{- 1} x + \frac{π}{2}$ . Hence find the greatest and least values of $f$ .

Try it first, then show the solutionSolution

$- \frac{π}{2} \leq sin^{- 1} x \leq \frac{π}{2}$ for $x \in [- 1, 1]$ . (½ mark)
Multiply by 2: $- π \leq 2 sin^{- 1} x \leq π$ . Add $\frac{π}{2}$ : $- \frac{π}{2} \leq f (x) \leq \frac{3 π}{2}$ . (1 mark)
Greatest value $\frac{3 π}{2}$ (at $x = 1$ ); least value $- \frac{π}{2}$ (at $x = - 1$ ). (½ mark)

Answer: Range $= [- \frac{π}{2}, \frac{3 π}{2}]$ ; greatest $\frac{3 π}{2}$ , least $- \frac{π}{2}$ .

Example 5: A double inequality that is half automatic · 3 marks · Short answerHOTS / competency

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

Try it first, then show the solutionSolution

Need $- 1 \leq x^{2} - 2 x \leq 1$ . (½ mark)
Left part: $x^{2} - 2 x + 1 \geq 0$ , that is $(x - 1)^{2} \geq 0$ , true for every real $x$ . (1 mark)
Right part: $x^{2} - 2 x - 1 \leq 0$ . The roots of $x^{2} - 2 x - 1 = 0$ are $1 \pm 2$ , so $1 - 2 \leq x \leq 1 + 2$ . (1 mark)
Intersecting with all reals leaves the second condition. (½ mark)

Answer: Domain $= [1 - 2, 1 + 2]$ .

Example 6: Case study: a ladder against a wall · 4 marks · Case-basedHOTS / competency

Suppose a 10 m ladder leans against a vertical wall with its foot $d$ m from the wall. The angle the ladder makes with the ground is $θ = cos^{- 1} (\frac{d}{10})$ .
(i) For which values of $d$ is $cos^{- 1} \frac{d}{10}$ defined? (1 mark)
(ii) Find $θ$ when $d = 5$ . (1 mark)
(iii) Safety rules need $\frac{π}{6} \leq θ \leq \frac{π}{3}$ . Find the corresponding range of $d$ . (2 marks)

Try it first, then show the solutionSolution

(i) $- 1 \leq \frac{d}{10} \leq 1 \Rightarrow - 10 \leq d \leq 10$ ; physically $0 \leq d \leq 10$ . (1 mark)
(ii) $θ = cos^{- 1} \frac{1}{2} = \frac{π}{3}$ . (1 mark)
(iii) $cos^{- 1}$ is decreasing, so a larger $θ$ needs a smaller $d$ . $θ = \frac{π}{3}$ gives $d = 10 cos \frac{π}{3} = 5$ ; $θ = \frac{π}{6}$ gives $d = 10 cos \frac{π}{6} = 53$ . (1 mark)
So $5 \leq d \leq 53$ (about 5 m to 8.66 m). (1 mark)

Answer: (i) $- 10 \leq d \leq 10$ (physically $0 \leq d \leq 10$ ). (ii) $\frac{π}{3}$ . (iii) $5 \leq d \leq 53$ .

Where students lose marks in Domain and Range of Inverse Trigonometric Functions

The slips examiners see most often, and what to write instead.

Mistake: Writing the range of tan inverse with square brackets
$tan \frac{π}{2}$ is undefined, so $\pm \frac{π}{2}$ are never values of $tan^{- 1} x$ .
Wrong working
Range of $tan^{- 1} x$ $= [- \frac{π}{2}, \frac{π}{2}]$ .
Right working
Range of $tan^{- 1} x$ $= (- \frac{π}{2}, \frac{π}{2})$ .
Do this: Use round brackets wherever the original trigonometric function is undefined.
Usually costs: The MCQ, or the final mark in a range question.
Mistake: Solving only half of a double inequality
With $x^{2}$ inside, each half gives a different set and both must hold.
Wrong working
$- 1 \leq x^{2} - 4 \leq 1 \Rightarrow x^{2} \leq 5 \Rightarrow - 5 \leq x \leq 5$ .
Right working
Also $x^{2} \geq 3$ , so the domain is $[- 5, - 3] \cup [3, 5]$ .
Do this: Split into 'left inequality' and 'right inequality', solve both, then intersect.
Usually costs: 1 to 2 marks.
Mistake: Using the sin-type range for cos inverse or sec inverse
Mixing up the two families gives negative angles for $cos^{- 1}$ .
Wrong working
$cos^{- 1} (- \frac{1}{2}) = - \frac{π}{3}$ .
Right working
$cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ , which lies in $[0, π]$ .
Do this: Remember: the cos family is never negative.
Usually costs: The whole value, and any later step that uses it.
Mistake: Taking the domain of sec inverse as all reals except ±1
The excluded set is the whole open interval $(- 1, 1)$ , not just two points.
Wrong working
Domain of $sec^{- 1} x$ $= R - {- 1, 1}$ .
Right working
Domain of $sec^{- 1} x$ $= R - (- 1, 1)$ , that is $x \leq - 1$ or $x \geq 1$ .
Do this: Picture the graph of $sec θ$ : its values are never strictly between $- 1$ and $1$ .
Usually costs: 1 mark.

Exam tips

Domain questions on inverse trigonometric functions appear almost every year as 1- or 2-mark items (2025 board paper, 2026 sample paper, 2027 sample paper); write the inequality you solve, not just the answer.
Keep the table of six domains and ranges ready; many MCQs on graphs simply test which interval the graph covers.
In a range question, say whether the function is increasing or decreasing before you order the end values.
For sums of inverse functions, find each domain separately and write the intersection as a final step.
Do not use identities like $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$ as a property; properties of inverse trigonometric functions are not in the 2026-27 syllabus.

Practice questions on Domain and Range of Inverse Trigonometric Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The range of $cos^{- 1} x$ is:
(a) $[- \frac{π}{2}, \frac{π}{2}]$
(b) $[0, π]$
(c) $(0, π)$
(d) $[0, π] - {\frac{π}{2}}$
Hint
The cos family runs from 0 to $π$ , and $cos$ is defined at both ends.
Answer
(b) $[0, π]$ .
Q2 · 1 mark · MCQBoard level
The domain of $sin^{- 1} (3 x)$ is:
(a) $[- 3, 3]$
(b) $[- 1, 1]$
(c) $[- \frac{1}{3}, \frac{1}{3}]$
(d) $R$
Hint
Solve $- 1 \leq 3 x \leq 1$ .
Answer
(c) $[- \frac{1}{3}, \frac{1}{3}]$ .
Q3 · 2 marks · Very short answerBoard level
Find the domain and range of $f (x) = cos^{- 1} (2 x - 3)$ .
Hint
The inside can take every value from $- 1$ to $1$ on the domain.
Answer
$- 1 \leq 2 x - 3 \leq 1 \Rightarrow 1 \leq x \leq 2$ , so the domain is $[1, 2]$ . As $x$ runs over $[1, 2]$ , $2 x - 3$ runs over $[- 1, 1]$ , so the range is $[0, π]$ .
Q4 · 2 marks · Very short answerHarder board level
Find the domain of $f (x) = sin^{- 1} x + sec^{- 1} x$ .
Hint
Intersect $[- 1, 1]$ with $(- \infty, - 1] \cup [1, \infty)$ .
Answer
$sin^{- 1} x$ needs $- 1 \leq x \leq 1$ ; $sec^{- 1} x$ needs $x \leq - 1$ or $x \geq 1$ . Both hold only at $x = - 1$ and $x = 1$ . Domain $= {- 1, 1}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The domain of $cosec^{- 1} x$ is $R - (- 1, 1)$ .
Reason (R): $∣ cosec θ ∣ \geq 1$ wherever $cosec θ$ is defined.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
$cosec θ = \frac{1}{s i n θ}$ and $∣ sin θ ∣ \leq 1$ .
Answer
(a). Since $0 < ∣ sin θ ∣ \leq 1$ where cosec is defined, $∣ cosec θ ∣ \geq 1$ ; these are exactly the values $cosec^{- 1}$ can accept, so R explains A.
Q6 · 2 marks · Very short answerHarder board level
Find the range of $cos^{- 1} x$ when $x$ is restricted to $[- \frac{1}{2}, \frac{3}{2}]$ .
Hint
$cos^{- 1}$ is decreasing, so the larger $x$ gives the smaller angle.
Answer
$cos^{- 1} \frac{3}{2} = \frac{π}{6}$ and $cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ . Since $cos^{- 1}$ is decreasing, the range is $[\frac{π}{6}, \frac{2 π}{3}]$ .
Q7 · 3 marks · Short answerHOTS / competency
Find the domain of $f (x) = cos^{- 1} x + \frac{1}{s i n ^{- 1} x}$ .
Hint
Besides $- 1 \leq x \leq 1$ , the square root in the denominator needs $sin^{- 1} x > 0$ .
Answer
$cos^{- 1} x$ and $sin^{- 1} x$ need $- 1 \leq x \leq 1$ . The term $\frac{1}{s i n ^{- 1} x}$ needs $sin^{- 1} x > 0$ , and since $sin^{- 1}$ is increasing with $sin^{- 1} 0 = 0$ , this means $x > 0$ . Domain $= (0, 1]$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the domain of $cos^{- 1} x$ ?Show answer
$[- 1, 1]$ .
2.What is the range of $cot^{- 1} x$ ?Show answer
$(0, π)$ .
3.What is the range of $cosec^{- 1} x$ ?Show answer
$[- \frac{π}{2}, \frac{π}{2}] - {0}$ .
4.What condition must $g (x)$ satisfy for $sec^{- 1} (g (x))$ to be defined?Show answer
$∣ g (x) ∣ \geq 1$ .
5.Find the domain of $sin^{- 1} (\frac{x}{2})$ .Show answer
$- 1 \leq \frac{x}{2} \leq 1$ , so $[- 2, 2]$ .
6.Is $\frac{π}{2}$ in the range of $sec^{- 1} x$ ?Show answer
No; $sec \frac{π}{2}$ is undefined, so $\frac{π}{2}$ is excluded.

Domain and Range of Inverse Trigonometric Functions: frequently asked questions

Why is the range of sin inverse [−π/2, π/2] and not [0, π]?

On [0, π] the sine function is not one-one: sin(π/6) = sin(5π/6). The interval [−π/2, π/2] is where sine is one-one and still takes every value from −1 to 1, so it is chosen as the principal value branch.

What is the domain of tan inverse x?

All real numbers. The tangent function takes every real value on (−π/2, π/2), so tan inverse accepts any real input, and its outputs stay strictly between −π/2 and π/2.

Which NCERT section gives the domains and ranges?

Section 2.2 (Basic Concepts) of Chapter 2, Inverse Trigonometric Functions, in NCERT Mathematics Part I for Class 12, which also draws their graphs.

Are graphs of inverse trigonometric functions in the 2026-27 syllabus?

Yes. The syllabus lists 'Graphs of inverse trigonometric functions'. Expect 1-mark MCQs that show a graph and ask which function or which branch it is.

Is the domain of sec inverse x the same as that of cosec inverse x?

Yes. Both are the real numbers outside the open interval (−1, 1), that is x ≤ −1 or x ≥ 1. Their ranges differ: [0, π] without π/2 for sec inverse, and [−π/2, π/2] without 0 for cosec inverse.