MathsCBSE Class 12 · Chapter 2

Inverse Trigonometric Functions Class 12: Notes, Principal Values and Solved Questions

Q: Are the properties of inverse trigonometric functions in the 2026-27 syllabus?

No. NCERT section 2.3 (properties) has been dropped, except the basic facts \sin(\sin^{-1}x) = x for x \in [-1, 1] and \sin^{-1}(\sin x) = x for x \in [-\frac{\pi}{2}, \frac{\pi}{2}] (and their analogues). Identities such as \tan^{-1}x + \tan^{-1}y are not required.

Q: How many marks does this chapter carry?

It shares Unit I with Relations and Functions; together they carry 8 of the 80 marks. In our question bank most questions from this chapter are 1-mark MCQs or 2-mark evaluations.

Q: Why is the principal range of cos⁻¹ [0, π] and not [−π/2, π/2]?

Cosine is not one-one on [-\frac{\pi}{2}, \frac{\pi}{2}] (it is even, so cos(−x) = cos x). On [0, π] it decreases from 1 to −1, taking every value once.

Q: Is $\sin^{-1}x$ the same as $\frac{1}{\sin x}$?

No. \sin^{-1}x is the angle whose sine is x; \frac{1}{\sin x} is \csc x.

Q: Can I use a calculator in the exam?

No. Principal values in board questions involve standard angles, so learn the values at 0, π/6, π/4, π/3, π/2 and their related angles.

Q: What happens to the range when the function is transformed, for example $2\sin^{-1}x + \pi$?

Apply the transformation to the principal range: \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}], so 2\sin^{-1}x \in [-\pi, \pi] and 2\sin^{-1}x + \pi \in [0, 2\pi].

Updated 14 October 2026 · Fact-checked

Inverse Trigonometric Functions is Chapter 2 of CBSE Class 12 Maths. It explains why trigonometric functions must be restricted to become one-one and onto, defines $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, $\cot^{-1}$, $\sec^{-1}$ and $\csc^{-1}$ with their domains and principal value branches, and draws their graphs. Board questions ask for principal values, domains, ranges and graph identification.

What Inverse Trigonometric Functions is about

Trigonometric functions repeat their values, so they are not one-one on R and have no inverse there. But if we restrict each one to a suitable interval on which it is one-one and onto its range, it becomes invertible. The inverse function, such as $sin^{- 1} x$ (read 'sine inverse x', also written arcsin x), returns an angle. The particular interval chosen gives the principal value branch, and the value of the inverse function in that branch is the principal value.

You must know, for all six functions, the domain, the range of the principal value branch and the graph (obtained by reflecting the restricted graph in the line y = x). With these you can evaluate expressions such as $sin^{- 1} (- \frac{1}{2})$ , $cos^{- 1} (cos \frac{7 π}{6})$ or $tan (tan^{- 1} (- 1) + \frac{π}{3})$ , and find domains such as that of $cos^{- 1} (3 x - 2)$ .

The NCERT properties of inverse trigonometric functions are not in the 2026-27 syllabus, except the basic facts $sin (sin^{- 1} x) = x$ for $x \in [- 1, 1]$ and $sin^{- 1} (sin x) = x$ for $x \in [- \frac{π}{2}, \frac{π}{2}]$ (and the same for the other functions on their branches). The chapter shares Unit I (8 marks) with Relations and Functions. In our question bank it appears in all eight board papers from 2023 to 2026, mostly as 1-mark MCQs and 2-mark evaluations.

Weightage: This chapter is part of Unit I, Relations and Functions, which carries 8 of the 80 theory marks together with Relations and Functions.Source: CBSE Mathematics syllabus 2026-27, Unit I Relations and Functions

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 3.5 marks per paper. See the questions

Key concepts

Why restrict the domain: principal value branch

NCERT section 2.1 and 2.2

A function has an inverse only if it is one-one and onto. $sin x$ on R is neither: it repeats every $2 π$ and its values lie only in [−1, 1]. Restricting sin to $[- \frac{π}{2}, \frac{π}{2}]$ and its codomain to [−1, 1] makes it a bijection, so it has an inverse

$sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$

Other intervals such as $[\frac{π}{2}, \frac{3 π}{2}]$ would also work and give other branches; the branch with range $[- \frac{π}{2}, \frac{π}{2}]$ is the principal value branch, and unless stated otherwise $sin^{- 1} x$ means the value in it.

Note: $sin^{- 1} x$ is not $(sin x)^{- 1} = \frac{1}{s i n x}$ . And $y = sin^{- 1} x$ means $sin y = x$ with y in the principal range.

Worked example

Explain why $sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$ and not $\frac{5 π}{6}$ , although $sin \frac{5 π}{6} = \frac{1}{2}$ too. [1 mark]

Try it first, then show the solutionSolution

The principal value must lie in $[- \frac{π}{2}, \frac{π}{2}]$ .
$\frac{π}{6}$ lies in this interval; $\frac{5 π}{6}$ does not.

Answer: Only $\frac{π}{6}$ is in the principal range $[- \frac{π}{2}, \frac{π}{2}]$ .

Domains and principal value branches of all six functions

NCERT section 2.2

Learn this table exactly:

$sin^{- 1}$ : domain $[- 1, 1]$ ; range $[- \frac{π}{2}, \frac{π}{2}]$
$cos^{- 1}$ : domain $[- 1, 1]$ ; range $[0, π]$
$csc^{- 1}$ : domain $R - (- 1, 1)$ ; range $[- \frac{π}{2}, \frac{π}{2}] - {0}$
$sec^{- 1}$ : domain $R - (- 1, 1)$ ; range $[0, π] - {\frac{π}{2}}$
$tan^{- 1}$ : domain R; range $(- \frac{π}{2}, \frac{π}{2})$
$cot^{- 1}$ : domain R; range $(0, π)$

Pattern: sin, cosec and tan use the interval around 0 (fourth and first quadrants); cos, sec and cot use $[0, π]$ (first and second quadrants). Open ends and removed points occur where the function is undefined: tan at $\pm \frac{π}{2}$ , cot at 0 and π, cosec at 0, sec at $\frac{π}{2}$ .

Worked example

Write the domain and range (principal value branch) of (a) $tan^{- 1} x$ , (b) $sec^{- 1} x$ . [2 marks]

Try it first, then show the solutionSolution

(a) $tan^{- 1}$ : domain R, range $(- \frac{π}{2}, \frac{π}{2})$
(b) $sec^{- 1}$ : domain $R - (- 1, 1)$ , i.e. $(- \infty, - 1] \cup [1, \infty)$ ; range $[0, π] - {\frac{π}{2}}$

Answer: (a) R, $(- \frac{π}{2}, \frac{π}{2})$ (b) $R - (- 1, 1)$ , $[0, π] - {\frac{π}{2}}$

Finding principal values, including negative arguments

NCERT section 2.2

To find a principal value, ask: which angle in the principal range has this trigonometric value?

$sin^{- 1} (- \frac{1}{2})$ : an angle in $[- \frac{π}{2}, \frac{π}{2}]$ with sine $- \frac{1}{2}$ is $- \frac{π}{6}$ .
$cos^{- 1} (- \frac{1}{2})$ : an angle in $[0, π]$ with cosine $- \frac{1}{2}$ is $\frac{2 π}{3}$ (not $- \frac{π}{3}$ , which is outside $[0, π]$ ).
$tan^{- 1} (- 1) = - \frac{π}{4}$ ; $cot^{- 1} (- 1) = \frac{3 π}{4}$ .
$sec^{- 1} (- 2) = \frac{2 π}{3}$ (cosine $- \frac{1}{2}$ in $[0, π]$ ); $csc^{- 1} (- 2) = - \frac{π}{6}$ .

So for sin, tan and cosec a negative argument gives a negative angle; for cos, cot and sec it gives an angle in the second quadrant. Values outside the domain (such as $sin^{- 1} 2$ or $sec^{- 1} \frac{1}{2}$ ) do not exist.

Worked example

Find the value of $cos^{- 1} (- \frac{3}{2}) + sin^{- 1} (- \frac{1}{2}) + tan^{- 1} (3)$ . [2 marks]

Try it first, then show the solutionSolution

$cos^{- 1} (- \frac{3}{2}) = \frac{5 π}{6}$ (in $[0, π]$ , $cos \frac{5 π}{6} = - \frac{3}{2}$ )
$sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ ; $tan^{- 1} 3 = \frac{π}{3}$
Sum $= \frac{5 π}{6} - \frac{π}{6} + \frac{π}{3} = \frac{4 π}{6} + \frac{2 π}{6} = π$

Answer: π

Composite expressions: f(f⁻¹ x) and f⁻¹(f x)

NCERT section 2.2

Two facts follow from the definition (and are the only 'properties' retained in the syllabus):

$sin (sin^{- 1} x) = x$ for $x \in [- 1, 1]$ ; similarly $cos (cos^{- 1} x) = x$ , $tan (tan^{- 1} x) = x$ for x in the domain.
$sin^{- 1} (sin x) = x$ only if $x \in [- \frac{π}{2}, \frac{π}{2}]$ ; similarly $cos^{- 1} (cos x) = x$ only for $x \in [0, π]$ , $tan^{- 1} (tan x) = x$ only for $x \in (- \frac{π}{2}, \frac{π}{2})$ .

If the angle is outside the principal range, first rewrite the inner value as the same trigonometric ratio of an angle that is inside: for example $sin \frac{3 π}{4} = sin \frac{π}{4}$ , so $sin^{- 1} (sin \frac{3 π}{4}) = \frac{π}{4}$ , not $\frac{3 π}{4}$ .

Worked example

Evaluate (a) $cos^{- 1} (cos \frac{7 π}{6})$ , (b) $tan^{- 1} (tan \frac{2 π}{3})$ . [2 marks]

Try it first, then show the solutionSolution

(a) $\frac{7 π}{6} \in / [0, π]$ . $cos \frac{7 π}{6} = cos (2 π - \frac{7 π}{6}) = cos \frac{5 π}{6}$ , and $\frac{5 π}{6} \in [0, π]$ , so the value is $\frac{5 π}{6}$ .
(b) $\frac{2 π}{3} \in / (- \frac{π}{2}, \frac{π}{2})$ . $tan \frac{2 π}{3} = tan (\frac{2 π}{3} - π) = tan (- \frac{π}{3})$ , so the value is $- \frac{π}{3}$ .

Answer: (a) $\frac{5 π}{6}$ (b) $- \frac{π}{3}$

Domain of composite inverse trigonometric functions

NCERT section 2.2

For $sin^{- 1} (g (x))$ or $cos^{- 1} (g (x))$ , the inside must lie in [−1, 1]: solve $- 1 \leq g (x) \leq 1$ . For $sec^{- 1} (g (x))$ or $csc^{- 1} (g (x))$ , solve $g (x) \leq - 1$ or $g (x) \geq 1$ . $tan^{- 1}$ and $cot^{- 1}$ accept all real inputs.

If the expression adds several functions, the domain is the intersection of their separate domains.

Worked example

Find the domain of (a) $cos^{- 1} (3 x - 2)$ , (b) $sin^{- 1} (x^{2} - 3)$ . [2 marks]

Try it first, then show the solutionSolution

(a) $- 1 \leq 3 x - 2 \leq 1 \Rightarrow 1 \leq 3 x \leq 3 \Rightarrow \frac{1}{3} \leq x \leq 1$
(b) $- 1 \leq x^{2} - 3 \leq 1 \Rightarrow 2 \leq x^{2} \leq 4 \Rightarrow 2 \leq ∣ x ∣ \leq 2$ , i.e. $x \in [- 2, - 2] \cup [2, 2]$

Answer: (a) $[\frac{1}{3}, 1]$ (b) $[- 2, - 2] \cup [2, 2]$

Graphs of inverse trigonometric functions

NCERT section 2.2

The graph of an inverse function is the mirror image of the graph of the (restricted) function in the line y = x: each point (a, b) becomes (b, a).

$y = sin^{- 1} x$ : from $(- 1, - \frac{π}{2})$ through the origin to $(1, \frac{π}{2})$ , increasing.
$y = cos^{- 1} x$ : from $(- 1, π)$ through $(0, \frac{π}{2})$ to $(1, 0)$ , decreasing.
$y = tan^{- 1} x$ : defined for all x, increasing, with horizontal asymptotes $y = \pm \frac{π}{2}$ .
$y = cot^{- 1} x$ : decreasing, asymptotes y = 0 and y = π.
$y = sec^{- 1} x$ and $y = csc^{- 1} x$ : two separate pieces, for $x \leq - 1$ and $x \geq 1$ .

Board MCQs often show a graph and ask which function it is: read the domain (x-extent) and range (y-extent) from the picture and match them with the table.

Worked example

A graph is defined only for $- 1 \leq x \leq 1$ , decreases from the point $(- 1, π)$ to the point $(1, 0)$ , and passes through $(0, \frac{π}{2})$ . Which function is it? [1 mark]

Try it first, then show the solutionSolution

Domain [−1, 1] means $sin^{- 1}$ or $cos^{- 1}$ .
Range $[0, π]$ and decreasing: $cos^{- 1} x$ .

Answer: $y = cos^{- 1} x$

Maths in real life: Inverse Trigonometric Functions

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.

Principal value of tan⁻¹

Suppose a ramp rises 1 m for every 1 m of horizontal distance, and an engineer wants its angle of slope with the horizontal.

Express the angle using an inverse trigonometric function and find it. Why is there only one sensible answer?

Try it first, then show the solutionSolution

Slope = rise/run = 1, so the angle θ satisfies $tan θ = 1$ .
$θ = tan^{- 1} 1 = \frac{π}{4} = 4 5^{\circ}$ , the principal value.
A slope angle lies between $- 9 0^{\circ}$ and $9 0^{\circ}$ , exactly the principal range $(- \frac{π}{2}, \frac{π}{2})$ of $tan^{- 1}$ .

Answer: $\frac{π}{4}$ (45°)

Memory hook: Slopes live in tan⁻¹'s range: between −90° and 90°.

Principal value of cos⁻¹

Imagine two forces pulling on a hook, and the cosine of the angle between them is found to be $- \frac{1}{2}$ .

Suppose the angle between two directions is θ with $cos θ = - \frac{1}{2}$ . Find θ, given that an angle between two directions lies between 0 and π.

Try it first, then show the solutionSolution

The angle between two directions lies in $[0, π]$ , the principal range of $cos^{- 1}$ .
$θ = cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$
That is 120°: the forces pull partly against each other.

Answer: $\frac{2 π}{3}$ (120°)

Memory hook: Angles between directions run from 0 to π, just like cos⁻¹.

Domain restriction

Suppose a calculator is asked for $sin^{- 1} (1.2)$ and shows an error message.

Explain the error. For what values of k does $sin^{- 1} (2 k - 1)$ make sense?

Try it first, then show the solutionSolution

Sine never exceeds 1 in size, so no angle has sine 1.2: 1.2 is outside the domain [−1, 1].
$- 1 \leq 2 k - 1 \leq 1 \Rightarrow 0 \leq 2 k \leq 2 \Rightarrow 0 \leq k \leq 1$

Answer: 1.2 is outside [−1, 1]; k must lie in [0, 1].

Memory hook: sin⁻¹ and cos⁻¹ only accept inputs between −1 and 1.

Topics in this chapter

Inverse Trigonometric Functions formula sheet

sin⁻¹: $sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$
cos⁻¹: $cos^{- 1} : [- 1, 1] \to [0, π]$
tan⁻¹: $tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2})$
cot⁻¹: $cot^{- 1} : R \to (0, π)$
sec⁻¹: $sec^{- 1} : R - (- 1, 1) \to [0, π] - {\frac{π}{2}}$
cosec⁻¹: $csc^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}] - {0}$
Meaning of the inverse: $y = sin^{- 1} x ⟺ sin y = x, y \in [- \frac{π}{2}, \frac{π}{2}]$
Function of its inverse: $sin (sin^{- 1} x) = x, x \in [- 1, 1]$
Inverse of the function: $sin^{- 1} (sin x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]$; Similarly $cos^{- 1} (cos x) = x$ for $x \in [0, π]$ , $tan^{- 1} (tan x) = x$ for $x \in (- \frac{π}{2}, \frac{π}{2})$ .

How the board asks Inverse Trigonometric Functions

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Principal values	1-2 marks: find $sin^{- 1} (- \frac{1}{2})$ , $cos^{- 1} (- \frac{3}{2})$ , $sec^{- 1} (- 2)$ , or a sum of several.	For each term find the angle in its principal range; take care with negative arguments for cos, sec and cot.
Inverse of a function outside its branch	Evaluate $sin^{- 1} (sin \frac{3 π}{4})$ , $cos^{- 1} (cos \frac{7 π}{6})$ , $tan^{- 1} (tan \frac{3 π}{4})$ , or such inside a sine or tangent.	Rewrite the inner ratio as the same ratio of an angle inside the principal range, then read off the angle.
Domains and ranges	Write domain and range of an inverse function; domain of $cos^{- 1} (3 x - 2)$ or $sin^{- 1} (x^{2} - 4)$ ; range of $2 sin^{- 1} x + c$ .	Use the table; for composites solve the inequality; for transformed functions apply the transformation to the range.
Graphs	MCQs with a graph to identify, or sketch a given inverse function.	Read x-extent (domain) and y-extent (range) and whether it rises or falls.
Assertion-reason	Statements about principal branches, domains or values that do not exist.	Check each statement against the table; note values outside the domain (such as $sec^{- 1} \frac{1}{2}$ ) are undefined.

Tricks to remember Inverse Trigonometric Functions

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

'Sin Tan Cosec' hug zero; 'Cos Cot Sec' start at zero
sin⁻¹, tan⁻¹, cosec⁻¹: ranges centred on 0, from $- \frac{π}{2}$ to $\frac{π}{2}$ . cos⁻¹, cot⁻¹, sec⁻¹: ranges from 0 to π.
Use it when: Writing principal branches.
Negative in, which quadrant out?
Negative argument: sin⁻¹, tan⁻¹, cosec⁻¹ give a negative angle (4th quadrant); cos⁻¹, cot⁻¹, sec⁻¹ give an obtuse angle (2nd quadrant).
Use it when: Principal values of negative numbers.
Holes where the parent is undefined
tan: open at ±π/2; cot: open at 0 and π; sec: missing π/2; cosec: missing 0.
Use it when: Exact range notation.
Mirror in y = x
Flip the restricted graph over y = x to get the inverse's graph; domain and range swap.
Use it when: Graph questions.
Inside first, then outside
For $sin^{- 1} (sin θ)$ , first move θ into the principal range keeping the same sine value, then cancel.
Use it when: Composite evaluations.

Inverse Trigonometric Functions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Principal value branch · 1 mark · MCQBoard level

The principal value branch of $sec^{- 1} x$ is:
(a) $[- \frac{π}{2}, \frac{π}{2}] - {0}$
(b) $[0, π] - {\frac{π}{2}}$
(c) $(0, π)$
(d) $(- \frac{π}{2}, \frac{π}{2})$

Try it first, then show the solutionSolution

sec belongs to the cos family: [0, π], with π/2 removed because sec is undefined there.

Answer: (b)

Example 2: Undefined value · 1 mark · Assertion-reasonBoard level

Assertion (A): $cos^{- 1} (2)$ is not defined.
Reason (R): The domain of $cos^{- 1}$ is [−1, 1].
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

No angle has cosine 2; 2 is outside [−1, 1]. R explains A.

Answer: (a)

Example 3: A sum of principal values · 2 marks · Very short answerHarder board level

Find the value of $2 cos^{- 1} (\frac{1}{2}) + 3 sin^{- 1} (- \frac{1}{2}) - tan^{- 1} (- \frac{1}{3})$ . [2 marks]

Try it first, then show the solutionSolution

$cos^{- 1} \frac{1}{2} = \frac{π}{3}$ ; $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ ; $tan^{- 1} (- \frac{1}{3}) = - \frac{π}{6}$ (each in its principal range)
Value $= \frac{2 π}{3} - \frac{3 π}{6} + \frac{π}{6} = \frac{4 π - 3 π + π}{6} = \frac{2 π}{6} = \frac{π}{3}$

Answer: $\frac{π}{3}$

Example 4: Angles outside the branch · 2 marks · Very short answerHarder board level

Evaluate $sin^{- 1} (sin \frac{5 π}{6}) + cos^{- 1} (cos \frac{13 π}{6})$ . [2 marks]

Try it first, then show the solutionSolution

$sin \frac{5 π}{6} = sin \frac{π}{6}$ , so the first term is $\frac{π}{6}$ .
$cos \frac{13 π}{6} = cos (2 π + \frac{π}{6}) = cos \frac{π}{6}$ , so the second term is $\frac{π}{6}$ . Sum $= \frac{π}{3}$ .

Answer: $\frac{π}{3}$

Example 5: Finding an unknown · 2 marks · Very short answerHarder board level

Find the value of k if $sin^{- 1} [k cos (cos^{- 1} \frac{1}{2})] = \frac{π}{6}$ . [2 marks]

Try it first, then show the solutionSolution

$cos (cos^{- 1} \frac{1}{2}) = \frac{1}{2}$ (since $\frac{1}{2} \in [- 1, 1]$ )
$sin^{- 1} \frac{k}{2} = \frac{π}{6} \Rightarrow \frac{k}{2} = sin \frac{π}{6} = \frac{1}{2} \Rightarrow k = 1$

Answer: k = 1

Example 6: Domain and range of a transformed function · 3 marks · Short answerHarder board level

Find (a) the domain of $f (x) = sin^{- 1} (2 x - 1) + cos^{- 1} x$ , (b) the range of $g (x) = 3 cos^{- 1} x - \frac{π}{2}$ . [3 marks]

Try it first, then show the solutionSolution

(a) $- 1 \leq 2 x - 1 \leq 1 \Rightarrow 0 \leq x \leq 1$ ; also $- 1 \leq x \leq 1$ for $cos^{- 1} x$ .
Intersection: $[0, 1]$ .
(b) $cos^{- 1} x \in [0, π] \Rightarrow 3 cos^{- 1} x \in [0, 3 π] \Rightarrow g (x) \in [- \frac{π}{2}, \frac{5 π}{2}]$ .

Answer: (a) [0, 1] (b) $[- \frac{π}{2}, \frac{5 π}{2}]$

Example 7: Sine of a sum of inverses · 2 marks · Very short answerHOTS / competency

Evaluate $sin [\frac{π}{3} - sin^{- 1} (- \frac{1}{2})]$ . [2 marks]

Try it first, then show the solutionSolution

$sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$
$sin [\frac{π}{3} + \frac{π}{6}] = sin \frac{π}{2} = 1$

Answer: 1

Example 8: Solving an inequality with cot⁻¹ · 2 marks · Very short answerHOTS / competency

Find the values of x for which $cot^{- 1} (x - 2) > \frac{π}{4}$ . [2 marks]

Try it first, then show the solutionSolution

$cot^{- 1}$ is decreasing on R with range $(0, π)$ , and $cot^{- 1} 1 = \frac{π}{4}$ .
So $cot^{- 1} (x - 2) > cot^{- 1} 1 ⟺ x - 2 < 1 ⟺ x < 3$ .

Answer: x < 3, i.e. $x \in (- \infty, 3)$

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

A street light and its shadow

Suppose a street lamp stands on a pole, and the angle θ of elevation of the lamp seen from a point on the road is given by $θ = tan^{- 1} (\frac{h}{d})$ , where h is the height of the lamp and d the horizontal distance of the point from the pole.

(i) · 1 mark
If h = d, find θ.
Answer
Answer: $\frac{π}{4}$
$tan^{- 1} 1 = \frac{π}{4}$
(ii) · 1 mark
If $h = 3 d$ , find θ.
Answer
Answer: $\frac{π}{3}$
$tan^{- 1} 3 = \frac{π}{3}$
(iii) · 2 marks
Write the domain and principal range of $tan^{- 1}$ , and explain why θ always lies between 0 and $\frac{π}{2}$ here.
Answer
Answer: Domain R, range $(- \frac{π}{2}, \frac{π}{2})$ ; h/d > 0, so θ is in $(0, \frac{π}{2})$ .
Positive arguments give angles between 0 and π/2 because tan⁻¹ is increasing with $tan^{- 1} 0 = 0$ .

Reading a graph

Suppose a student is shown the graph of a function that exists for every real x, increases throughout, passes through the origin, and approaches the lines $y = \frac{π}{2}$ and $y = - \frac{π}{2}$ without touching them.

(i) · 1 mark
Name the function.
Answer
Answer: $y = tan^{- 1} x$
Domain R, range $(- \frac{π}{2}, \frac{π}{2})$ , increasing.
(ii) · 1 mark
What is its value at x = −1?
Answer
Answer: $- \frac{π}{4}$
Angle in the range with tangent −1.
(iii) · 2 marks
Sketch, in words, the graph of $y = cot^{- 1} x$ and give its range.
Answer
Answer: Decreasing curve for all real x, approaching y = π as x → −∞ and y = 0 as x → +∞, passing through $(0, \frac{π}{2})$ ; range $(0, π)$ .
Reflect the graph of cot on $(0, π)$ in y = x.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Evaluate $tan (tan^{- 1} (- 1) + \frac{π}{3})$ . [2 marks] (2 marks)

Step	Marks
$tan^{- 1} (- 1) = - \frac{π}{4}$ , so the angle is $- \frac{π}{4} + \frac{π}{3} = \frac{π}{12}$	1
$tan \frac{π}{12} = tan (4 5^{\circ} - 3 0^{\circ}) = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{3 - 1}{3 + 1} = 2 - 3$	1

Find the value of $sin^{- 1} (sin \frac{3 π}{5}) + cos^{- 1} (cos \frac{7 π}{6})$ . [3 marks] (3 marks)

Step	Marks
$sin \frac{3 π}{5} = sin (π - \frac{3 π}{5}) = sin \frac{2 π}{5}$ , with $\frac{2 π}{5} \in [- \frac{π}{2}, \frac{π}{2}]$ , so the first term is $\frac{2 π}{5}$	1
$cos \frac{7 π}{6} = cos \frac{5 π}{6}$ , with $\frac{5 π}{6} \in [0, π]$ , so the second term is $\frac{5 π}{6}$	1
Sum $= \frac{2 π}{5} + \frac{5 π}{6} = \frac{12 π + 25 π}{30} = \frac{37 π}{30}$	1

(a) Write the domain and principal value branch of $sin^{- 1} x$ and $cos^{- 1} x$ , and sketch their graphs. (b) Find the domain of $cos^{- 1} (3 x - 2)$ . (c) Find the principal value of $cos^{- 1} (- \frac{1}{2}) - 2 sin^{- 1} (- \frac{1}{2})$ . (A combined practice question in 5-mark style.) [5 marks] (5 marks)

Step	Marks
(a) $sin^{- 1}$ : [−1, 1] → $[- \frac{π}{2}, \frac{π}{2}]$ ; $cos^{- 1}$ : [−1, 1] → $[0, π]$	1
Graphs: $sin^{- 1}$ increasing from $(- 1, - \frac{π}{2})$ to $(1, \frac{π}{2})$ ; $cos^{- 1}$ decreasing from $(- 1, π)$ to $(1, 0)$	1
(b) $- 1 \leq 3 x - 2 \leq 1 \Rightarrow \frac{1}{3} \leq x \leq 1$	1
(c) $cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ and $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$	1
$\frac{2 π}{3} - 2 (- \frac{π}{6}) = \frac{2 π}{3} + \frac{π}{3} = π$	1

Where students lose marks in Inverse Trigonometric Functions

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

Mistake: Writing $cos^{- 1} (- \frac{1}{2}) = - \frac{π}{3}$
Copying the sin⁻¹ pattern.
Wrong working
$cos^{- 1} (- \frac{1}{2}) = - \frac{π}{3}$
Right working
$cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ (must lie in $[0, π]$ )
Do this: Check the answer lies in the principal range of that particular function.
Usually costs: 1 mark per term.
Mistake: Cancelling $sin^{- 1} (sin x) = x$ for any x
Treating the functions as exact inverses everywhere.
Wrong working
$sin^{- 1} (sin \frac{3 π}{4}) = \frac{3 π}{4}$
Right working
$= sin^{- 1} (sin \frac{π}{4}) = \frac{π}{4}$
Do this: Cancel only when x is in the principal range.
Usually costs: Full marks for that part.
Mistake: Writing $sin^{- 1} x$ as $\frac{1}{s i n x}$
Misreading the notation.
Do this: $sin^{- 1} x$ is an angle; $(sin x)^{- 1} = csc x$ .
Usually costs: 1 mark.
Mistake: Using square brackets where the range is open
Careless notation.
Do this: tan⁻¹ range is $(- \frac{π}{2}, \frac{π}{2})$ , open; cot⁻¹ is $(0, π)$ , open.
Usually costs: ½ mark.
Mistake: Using dropped identities such as $tan^{- 1} x + tan^{- 1} y$
Older syllabi.
Do this: These properties are not in the 2026-27 syllabus; solve questions from the definitions and principal ranges.
Usually costs: Time, and risk of wrong methods.

NCERT exercise guide: Inverse Trigonometric Functions

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 2.1 (principal values)

What it tests: Principal values of all six inverse functions at standard values, sums of principal values, and MCQs on principal branches.

How to approach it: For each term find the angle in its principal range; add carefully with common denominators.

Same kind as Exercise 2.1

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

Show the solutionSolution

$cot^{- 1} (- 3)$ : angle in $(0, π)$ with cotangent $- 3$ is $\frac{5 π}{6}$ .
$sec^{- 1} \frac{2}{3}$ : cosine $\frac{3}{2}$ in $[0, π]$ gives $\frac{π}{6}$ .
Sum $= \frac{5 π}{6} + \frac{π}{6} = π$

Answer: π

Miscellaneous questions on $f^{- 1} (f (x))$ (retained part)

What it tests: Values such as $sin^{- 1} (sin \frac{2 π}{3})$ and $tan^{- 1} (tan \frac{3 π}{4})$ using only the definition of the principal branch.

How to approach it: Bring the angle into the principal range with the same trigonometric value, then read it off.

Same kind as the miscellaneous exercise

Find $tan^{- 1} (tan \frac{3 π}{4})$ .

Show the solutionSolution

$\frac{3 π}{4} \in / (- \frac{π}{2}, \frac{π}{2})$
$tan \frac{3 π}{4} = tan (\frac{3 π}{4} - π) = tan (- \frac{π}{4})$
So the value is $- \frac{π}{4}$

Answer: $- \frac{π}{4}$

Inverse Trigonometric Functions: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 3.5 marks per paper; 18 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Inverse Trigonometric Functions

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 65/1/1, Q1 · 1 mark · MCQ

If $2 cos^{- 1} x = y$ , then

(A) $0 \leq y \leq π$
(B) $- π \leq y \leq π$
(C) $0 \leq y \leq 2 π$
(D) $- π \leq y \leq 0$

Try it first, then show our solutionSolution

Principal range of $cos^{- 1} x$ is $[0, π]$ , so $0 \leq cos^{- 1} x \leq π \Rightarrow 0 \leq 2 cos^{- 1} x \leq 2 π$ , i.e. $0 \leq y \leq 2 π$ . Option (C).
1 mark

Answer: $0 \leq y \leq 2 π$ , option (C)

Where the marks are

Uses range $[0, π]$ of $cos^{- 1} x$ and doubles it (1 mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/1/1, Q25(a) · 2 marks · Very short answer

Simplify : $tan^{- 1} (\frac{c o s 2 x - s i n 2 x}{c o s 2 x + s i n 2 x})$ , $0 < x < \frac{π}{4}$ .

Try it first, then show our solutionSolution

Divide numerator and denominator by $cos 2 x$ ( $\neq = 0$ since $0 < 2 x < \frac{π}{2}$ ):
$tan^{- 1} (\frac{1 - tan 2 x}{1 + tan 2 x})$
1 mark
$= tan^{- 1} [tan (\frac{π}{4} - 2 x)] = \frac{π}{4} - 2 x$ , since $- \frac{π}{4} < \frac{π}{4} - 2 x < \frac{π}{4}$ lies in $(- \frac{π}{2}, \frac{π}{2})$ .
1 mark

Answer: $\frac{π}{4} - 2 x$

Where the marks are

Converts to $\frac{1 - t a n 2 x}{1 + t a n 2 x}$ (1 mark)
Writes as $tan (\frac{π}{4} - 2 x)$ and simplifies to $\frac{π}{4} - 2 x$ (1 mark)

See it in the full 2026 board paper, set 65/1/1

2024, set 65/1/1, Q38 · 4 marks · Case-basedCompetency-based

If a function $f : X \to Y$ defined as $f (x) = y$ is one-one and onto, then we can define a unique function $g : Y \to X$ such that $g (y) = x$ , where $x \in X$ and $y = f (x), y \in Y$ . Function g is called the inverse of function f.

The domain of sine function is R and function sine : $R \to R$ is neither one-one nor onto. The following graph shows the sine function.

Let sine function be defined from set A to $[- 1, 1]$ such that inverse of sine function exists, i.e., $sin^{- 1} x$ is defined from $[- 1, 1]$ to A.

On the basis of the above information, answer the following questions :

(i)
If A is the interval other than principal value branch, give an example of one such interval.
1 mark
(ii)
If $sin^{- 1} (x)$ is defined from $[- 1, 1]$ to its principal value branch, find the value of $sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1)$ .
1 mark
(iii)
(a) Draw the graph of $sin^{- 1} x$ from $[- 1, 1]$ to its principal value branch.
2 marks
OR
(iii)
(b) Find the domain and range of $f (x) = 2 sin^{- 1} (1 - x)$ .
2 marks

Figure in the paper: Graph of y = sin x on X'X and Y'Y axes, a wave passing through the origin, with marked x-values at -5π/2, -2π, -π, -π/2, π/2, π, 3π/2, 2π, 5π/2 and y-values 1 and -1. See it in the official paper.

Try it first, then show our solutionSolution

(i) On $[\frac{π}{2}, \frac{3 π}{2}]$ , sine is one-one and takes every value in $[- 1, 1]$ (it falls from 1 to −1). So $A = [\frac{π}{2}, \frac{3 π}{2}]$ works. Any such interval, e.g. $[- \frac{3 π}{2}, - \frac{π}{2}]$ , is also correct.
1 mark
(ii) With principal values in $[- \frac{π}{2}, \frac{π}{2}]$ :
$sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1) = - \frac{π}{6} - \frac{π}{2} = - \frac{2 π}{3}$
1 mark
(iii)(a) **Graph of $y = sin^{- 1} x$ , domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ :**
- Plot the points $(- 1, - \frac{π}{2})$ , $(- \frac{1}{2}, - \frac{π}{6})$ , $(0, 0)$ , $(\frac{1}{2}, \frac{π}{6})$ , $(1, \frac{π}{2})$ .
- Join them with a smooth increasing curve through the origin. It is symmetric about the origin and becomes vertical at the end-points $x = \pm 1$ .
- This curve is the mirror image of $y = sin x$ , $x \in [- \frac{π}{2}, \frac{π}{2}]$ , in the line $y = x$ .
- Mark $\pm \frac{π}{2}$ on the y-axis and $\pm 1$ on the x-axis.
2 marks
(iii)(b) Domain: $sin^{- 1} u$ is defined only for $- 1 \leq u \leq 1$ , so $- 1 \leq 1 - x \leq 1 \Rightarrow - 2 \leq - x \leq 0 \Rightarrow 0 \leq x \leq 2$ .
½ mark
(iii)(b) Domain $= [0, 2]$ .
½ mark
(iii)(b) Range: $- \frac{π}{2} \leq sin^{- 1} (1 - x) \leq \frac{π}{2}$ ; multiplying by 2, $- π \leq 2 sin^{- 1} (1 - x) \leq π$ .
½ mark
(iii)(b) Range $= [- π, π]$ .
½ mark

Answer: (i) $[\frac{π}{2}, \frac{3 π}{2}]$ (or any similar interval); (ii) $- \frac{2 π}{3}$ ; (iii)(a) increasing curve from $(- 1, - π /2)$ through $(0, 0)$ to $(1, π /2)$ ; (iii)(b) domain $[0, 2]$ , range $[- π, π]$

Where the marks are

(i) Gives a valid interval other than the principal branch, e.g. $[π /2, 3 π /2]$ (1 mark)
(ii) Uses the principal values $- π /6$ and $π /2$ to get $- 2 π /3$ (1 mark)
(iii)(a) Correct graph of $sin^{- 1} x$ on $[- 1, 1]$ with range $[- π /2, π /2]$ marked (2 marks)
(iii)(b) Writes the condition $- 1 \leq 1 - x \leq 1$ (½ mark)
(iii)(b) Gets domain $[0, 2]$ (½ mark)
(iii)(b) Uses the bounds $- π /2 \leq sin^{- 1} (1 - x) \leq π /2$ (½ mark)
(iii)(b) Gets range $[- π, π]$ (½ mark)

See it in the full 2024 board paper, set 65/1/1

Every Inverse Trigonometric Functions question, by year

For 2027

Sample paper 2026-27: Q1 (1 mark, solved), Q2 (1 mark, solved), Q3 (1 mark, solved)

2026

Set 65/1/1: Q1 (1 mark, solved), Q25(a) (2 marks, solved), Q25(b) (2 marks, solved)

Set 65/5/1: Q3 (1 mark, solved), Q21 (2 marks, solved)

Sample paper 2025-26: Q1 (1 mark, solved), Q19 (1 mark, solved), Q21(A) (2 marks, solved), Q21(B) (2 marks, solved)

2025

Set 65/1/1: Q18 (1 mark, solved), Q22 (2 marks, solved)

Set 65/7/1: Q1 (1 mark, solved), Q2 (1 mark, solved), Q20 (1 mark, solved)

Sample paper 2024-25: Q15 (1 mark, solved), Q21 (2 marks, solved)

2024

Set 65/1/1: Q38 (4 marks, solved)

Set 65/5/1: Q19 (1 mark, solved), Q21 (2 marks, solved)

Sample paper 2023-24: Q21(a) (2 marks, solved), Q21(b) (2 marks, solved)

2023

Set 65/1/1: Q19 (1 mark, solved), Q21(b) (2 marks, solved)

Set 65/5/1: Q2 (1 mark, solved), Q21 (2 marks, solved), Q27(a) (3 marks, solved)

Practice questions on 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 principal value of $sin^{- 1} (- \frac{3}{2})$ is:
(a) $\frac{4 π}{3}$
(b) $- \frac{π}{3}$
(c) $\frac{2 π}{3}$
(d) $\frac{5 π}{3}$
Hint
Must be in $[- \frac{π}{2}, \frac{π}{2}]$ .
Answer
(b) $- \frac{π}{3}$
Q2 · 1 mark · MCQBoard level
The domain of $cos^{- 1} (2 x)$ is:
(a) [−1, 1]
(b) [−2, 2]
(c) $[- \frac{1}{2}, \frac{1}{2}]$
(d) [0, 1]
Hint
$- 1 \leq 2 x \leq 1$ .
Answer
(c)
Q3 · 1 mark · Very short answerBoard level
Write the principal value of $cot^{- 1} (- 1)$ .
Hint
Range $(0, π)$ .
Answer
$\frac{3 π}{4}$
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $tan^{- 1} (tan \frac{3 π}{4}) = \frac{3 π}{4}$ .
Reason (R): $tan^{- 1} (tan x) = x$ for $x \in (- \frac{π}{2}, \frac{π}{2})$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Is 3π/4 in the range?
Answer
(d). A is false (the value is $- \frac{π}{4}$ ); R is true.
Q5 · 2 marks · Very short answerHarder board level
Find the value of $sec^{- 1} (- 2) + csc^{- 1} (- 2)$ .
Hint
Different ranges.
Answer
$\frac{2 π}{3} + (- \frac{π}{6}) = \frac{π}{2}$
Q6 · 2 marks · Very short answerHarder board level
Evaluate $cos [cos^{- 1} (- \frac{3}{2}) + \frac{π}{6}]$ .
Hint
Find the inner angle first.
Answer
$cos^{- 1} (- \frac{3}{2}) = \frac{5 π}{6}$ ; $cos (\frac{5 π}{6} + \frac{π}{6}) = cos π = - 1$ .
Q7 · 2 marks · Very short answerHarder board level
Find the domain of $f (x) = sin^{- 1} (x^{2} - 4)$ .
Hint
$- 1 \leq x^{2} - 4 \leq 1$ .
Answer
$3 \leq x^{2} \leq 5$ : $x \in [- 5, - 3] \cup [3, 5]$ .
Q8 · 3 marks · Short answerHOTS / competency
Evaluate $sin^{- 1} (cos \frac{28 π}{5})$ .
Hint
Reduce the angle, then convert cos to sin.
Answer
$\frac{28 π}{5} = 4 π + \frac{8 π}{5}$ , so $cos \frac{28 π}{5} = cos \frac{8 π}{5} = cos (2 π - \frac{8 π}{5}) = cos \frac{2 π}{5} = sin (\frac{π}{2} - \frac{2 π}{5}) = sin \frac{π}{10}$ . Since $\frac{π}{10}$ is in the principal range, the value is $\frac{π}{10}$ .
Q9 · 2 marks · Very short answerHarder board level
Find the range of $f (x) = 2 sin^{- 1} x + \frac{π}{2}$ , $x \in [- 1, 1]$ . What is the greatest value of f and where does it occur?
Hint
Transform the range.
Answer
$2 sin^{- 1} x \in [- π, π]$ , so $f \in [- \frac{π}{2}, \frac{3 π}{2}]$ . Greatest value $\frac{3 π}{2}$ at x = 1.
Q10 · 2 marks · Very short answerHOTS / competency
If $sin^{- 1} x = y$ , find the possible values of y, and hence of $cos y$ in terms of x.
Hint
y is in the principal range, where cos y ≥ 0.
Answer
$y \in [- \frac{π}{2}, \frac{π}{2}]$ , so $cos y \geq 0$ ; $sin y = x$ gives $cos y = 1 - x^{2}$ .
Q11 · 3 marks · Short answerHarder board level
Sketch the graphs of $y = sin^{- 1} x$ and $y = cos^{- 1} x$ on the same axes and find the x-coordinate of their point of intersection.
Hint
Equal angles where sin y = cos y.
Answer
sin⁻¹x rises from $(- 1, - \frac{π}{2})$ to $(1, \frac{π}{2})$ ; cos⁻¹x falls from $(- 1, π)$ to $(1, 0)$ . They meet where $sin^{- 1} x = cos^{- 1} x = y$ , so $sin y = cos y$ with $y \in [0, \frac{π}{2}]$ : $y = \frac{π}{4}$ , $x = \frac{1}{2}$ .

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.Principal range of $sin^{- 1}$ ?Show answer
$[- \frac{π}{2}, \frac{π}{2}]$ .
2.Principal range of $cot^{- 1}$ ?Show answer
$(0, π)$ .
3.Domain of $sec^{- 1}$ ?Show answer
$R - (- 1, 1)$ .
4.Value of $tan^{- 1} (- 3)$ ?Show answer
$- \frac{π}{3}$ .
5.Value of $cos^{- 1} (- 1)$ ?Show answer
π.
6.Value of $sin^{- 1} (sin \frac{2 π}{3})$ ?Show answer
$\frac{π}{3}$ .
7.Domain of $sin^{- 1} (2 x)$ ?Show answer
$[- \frac{1}{2}, \frac{1}{2}]$ .
8.Value of $csc^{- 1} (- 2)$ ?Show answer
$- \frac{π}{4}$ .

Best order to study it

Revise the values of sin, cos and tan at standard angles in all four quadrants.
Understand why a function must be one-one and onto to have an inverse, and why we restrict trigonometric functions.
Learn the domain and principal range table for all six inverse functions.
Practise principal values, especially with negative arguments.
Practise $f^{- 1} (f (x))$ evaluations for angles outside the principal range, and domains of composites.
Finish with graphs: sketch each one and identify graphs from their domain and range.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the domain and principal range of all six inverse functions from memory?
Can you find $sin^{- 1}$ , $cos^{- 1}$ and $tan^{- 1}$ of standard negative values?
Can you find $sec^{- 1} (- 2)$ and $csc^{- 1} (- 2)$ ?
Can you evaluate $sin^{- 1} (sin θ)$ and $cos^{- 1} (cos θ)$ for θ outside the principal range?
Can you evaluate expressions like $tan (tan^{- 1} (- 1) + \frac{π}{3})$ ?
Can you find the domain of $cos^{- 1} (a x + b)$ and $sin^{- 1} (x^{2} - c)$ ?
Can you find the range of $a sin^{- 1} x + b$ ?
Can you sketch the graphs of $sin^{- 1} x$ , $cos^{- 1} x$ and $tan^{- 1} x$ ?
Can you identify an inverse function from its graph?
Can you explain why $sin^{- 1} 2$ does not exist?

Inverse Trigonometric Functions: frequently asked questions

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

No. NCERT section 2.3 (properties) has been dropped, except the basic facts \sin(\sin^{-1}x) = x for x \in [-1, 1] and \sin^{-1}(\sin x) = x for x \in [-\frac{\pi}{2}, \frac{\pi}{2}] (and their analogues). Identities such as \tan^{-1}x + \tan^{-1}y are not required.

How many marks does this chapter carry?

It shares Unit I with Relations and Functions; together they carry 8 of the 80 marks. In our question bank most questions from this chapter are 1-mark MCQs or 2-mark evaluations.

Why is the principal range of cos⁻¹ [0, π] and not [−π/2, π/2]?

Cosine is not one-one on [-\frac{\pi}{2}, \frac{\pi}{2}] (it is even, so cos(−x) = cos x). On [0, π] it decreases from 1 to −1, taking every value once.

Is $\sin^{-1}x$ the same as $\frac{1}{\sin x}$?

No. \sin^{-1}x is the angle whose sine is x; \frac{1}{\sin x} is \csc x.

Can I use a calculator in the exam?

No. Principal values in board questions involve standard angles, so learn the values at 0, π/6, π/4, π/3, π/2 and their related angles.

What happens to the range when the function is transformed, for example $2\sin^{-1}x + \pi$?

Apply the transformation to the principal range: \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}], so 2\sin^{-1}x \in [-\pi, \pi] and 2\sin^{-1}x + \pi \in [0, 2\pi].

Based on: CBSE Mathematics (041) syllabus 2026-27; NCERT Mathematics Part I, Textbook for Class XII, Chapter 2 (Reprint 2026-27)

Inverse Trigonometric Functions Class 12: Notes, Principal Values and Solved Questions

What Inverse Trigonometric Functions is about

Key concepts

Maths in real life: Inverse Trigonometric Functions

Topics in this chapter

Inverse Trigonometric Functions formula sheet

How the board asks Inverse Trigonometric Functions

Tricks to remember Inverse Trigonometric Functions

Inverse Trigonometric Functions: board-level solved questions

Case-study practice (Section E style)

A street light and its shadow

Reading a graph

Where the marks are

Where students lose marks in Inverse Trigonometric Functions

NCERT exercise guide: Inverse Trigonometric Functions

Exercise 2.1 (principal values)

Miscellaneous questions on f−1(f(x)) (retained part)

Inverse Trigonometric Functions: previous year questions

Solved board questions from Inverse Trigonometric Functions

Every Inverse Trigonometric Functions question, by year

Practice questions on Inverse Trigonometric Functions

Test your recall

Best order to study it

Quick revision checklist

Inverse Trigonometric Functions: frequently asked questions

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

How many marks does this chapter carry?

Why is the principal range of cos⁻¹ [0, π] and not [−π/2, π/2]?

Is $\sin^{-1}x$ the same as $\frac{1}{\sin x}$?

Can I use a calculator in the exam?

What happens to the range when the function is transformed, for example $2\sin^{-1}x + \pi$?

Miscellaneous questions on $f^{- 1} (f (x))$ (retained part)