State True or False: The domain of trigonometric functions can be restricted to any one of their branch (not n

NCERT Exemplar · Q51

Q. State True or False: The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

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Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

Take $\sin x=\tfrac12$ . It has infinitely many solutions: $x=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},-\tfrac{7\pi}{6},\dots$ — every angle whose sine is $\tfrac12$ . So if we want an inverse that returns a single angle for $\sin^{-1}(0.5)$ , we must first agree on one angle to report. A function is allowed only one output per input, and $\sin x$ over all of $\mathbb{R}$ is many-to-one — it fails the horizontal line test and cannot be inverted as it stands.

The idea: restrict to one clean interval

For each trigonometric ratio we restrict the angle to a single standard interval on which the function is one-to-one while still covering its entire range exactly once. On that interval the inverse becomes well-defined and single-valued. That interval — the set of angles the inverse is allowed to return — is the principal value branch (also called the principal value domain).

The interval is chosen to be strictly monotonic, to hit every output once, and to sit as close to $0$ as possible. For sine that is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ , where $\sin$ increases from $-1$ to $1$ .

The principal value branch of an inverse trig function is the interval of angles it returns — the restricted interval on which the original ratio is one-to-one and onto its range.

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

Why the intervals differ …

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The statement is True. Inverse trigonometric functions are defined by restricting the domain of the original trigonometric function to any interval where it is one‑one and onto. The principal value branch is just one such convenient choice — any branch that makes the function bijective works.

Why this question matters

The key idea is that a function must be one‑one (injective) and onto (surjective) to have an inverse. Trigonometric functions like $\sin x$ , $\cos x$ , $\sec x$ are periodic — they repeat values infinitely many times. So over their natural domain $\mathbb{R}$ , they are not one‑one. To define an inverse, we must restrict the domain to a smaller interval where the function becomes bijective.

The principal value branch is the most commonly used restriction (e.g., $\sin^{-1} x$ uses $[-\pi/2, \pi/2]$ ). But it is not the only possible one. Any interval that gives a one‑one and onto mapping will work — it just yields a different "branch" of the inverse function.

Step‑by‑step reasoning

What does "branch" mean here?

A branch of a trigonometric function is a continuous piece of its graph over an interval where the function is strictly monotonic (and hence one‑one). For example, $\sin x$ is strictly increasing on $[-\pi/2, \pi/2]$ , strictly decreasing on $[\pi/2, 3\pi/2]$ , and so on. Each such interval is a branch.
The condition for an inverse to exist

For $f: A \to B$ to have an inverse $f^{-1}: B \to A$ , $f$ must be bijective (one‑one and onto).
- One‑one: No two different $x$ in $A$ give the same $y$ .
- Onto: Every $y$ in $B$ is the image of some $x$ in $A$ .
For $\sin x$ , if we take $A = [-\pi/2, \pi/2]$ and $B = [-1, 1]$ , $\sin$ is bijective. That's the principal branch. But if we take $A = [\pi/2, 3\pi/2]$ , $\sin$ is also bijective onto $[-1, 1]$ (it's decreasing, but still one‑one and onto). So that's another valid branch.
Why the principal branch is "principal" but not unique

The principal value branch is chosen by convention — it's the one that gives the simplest range (usually centred at 0). But mathematically, any branch that makes the function bijective is equally valid. For instance:
- $\sec^{-1} x$ is usually defined with range $[0, \pi] \setminus \{\pi/2\}$ , but $[\pi, 2\pi] \setminus \{3\pi/2\}$ would also work.
- $\tan^{-1} x$ uses $(-\pi/2, \pi/2)$ , but $(\pi/2, 3\pi/2)$ is another branch.
The statement in the question …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 66 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$2\cos^{-1}x = y$
›Reveal solutionSolution
The range of $\cos^{-1}x$ is $[0,\pi]$ , so multiplying by 2 gives $y = 2\cos^{-1}x$ a range of $[0,2\pi]$ . The correct option is (C).

Concept and Intuition

The key to this problem lies entirely in understanding the range of the inverse cosine function. $\cos^{-1}x$ (also written as $\arccos x$ ) is defined as the angle whose cosine is $x$ , and by convention, that angle is always taken from the interval $[0, \pi]$ . This is not arbitrary — it's the standard principal value branch that makes the function one-to-one and therefore invertible.

Once you know that $\cos^{-1}x$ lives between $0$ and $\pi$ (inclusive), finding the range of $y = 2\cos^{-1}x$ is simply a matter of scaling that interval by a factor of 2. No tricky domain restrictions, no sign flips — just multiplication.

Watch out

A common mistake is to confuse the range of $\cos^{-1}x$ with that of $\sin^{-1}x$ (which is $[-\pi/2, \pi/2]$ ). Always recall: $\cos^{-1}x \in [0,\pi]$ , not $[-\pi/2,\pi/2]$ .

Step-by-step solution

Recall the range of $\cos^{-1}x$ The inverse cosine function $\cos^{-1}: [-1,1] \to [0,\pi]$ gives an output angle in radians. This means:

$0 \le \cos^{-1}x \le \pi \quad \text{for all } x \in [-1,1].$

Multiply the inequality by 2 Since 2 is positive, multiplying through preserves the direction of the inequalities:

$2 \cdot 0 \le 2\cos^{-1}x \le 2 \cdot \pi$
which simplifies to:
$0 \le y \le 2\pi.$

Check if every value in $[0,2\pi]$ is actually attained …
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CBSE 2026Set V11 markMCQ
$\tan^{-1} x$
(a) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(b) $(0, \pi)$
(c) $[-1, 1]$
(d) $(-\infty, \infty)$
›Reveal solutionSolution

The tangent function maps $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ onto all of $\mathbb{R}$ , so $\tan^{-1}x$ accepts every real $x$ ; answer (d).

The principal-branch tangent $\tan:\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\to\mathbb{R}$ is a bijection onto $\mathbb{R}$ . Its inverse $\tan^{-1}$ therefore has domain equal to the range of $\tan$ , namely all real …
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CBSE 2026Set CX1 mark
$\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$
›Reveal solutionSolution

$\tan^{-1}\sqrt3=\tfrac{\pi}{3}$ , $\sec^{-1}(-2)=\tfrac{2\pi}{3}$ , giving $-\tfrac{\pi}{3}$ .

Concept: Use the principal-value ranges: $\tan^{-1}\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $\sec^{-1}\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ .

$\tan^{-1}\sqrt3=\dfrac{\pi}{3}\qquad(\tan\tfrac{\pi}{3}=\sqrt3).$ …

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CBSE 2026Set ANNUAL1 mark
$\sin^{-1}x$
›Reveal solutionSolution

$\sin^{-1}x$ is defined only where $\sin\theta=x$ has a solution, i.e. for $x\in[-1,1]$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$y = \cos^{-1} x$
(a) $0 \le y \le \pi$
(b) $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
(c) $-\pi \le y \le \pi$
(d) None of these
›Reveal solutionSolution

$\cos^{-1}x$ is defined so that its principal value always lies in $[0,\pi]$ .

The function $\cos x$ is one-one and onto from $[0,\pi]$ to $[-1,1]$ , so its inverse $\cos^{-1}x$ is defined on domain $[-1,1]$ with range (principal value …

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CBSE 2026Set ANNUAL1 markMCQ
$\tan^{-1}(-1)$
(a) $\dfrac{\pi}{4}$
(b) $-\dfrac{\pi}{4}$
(c) $\dfrac{3\pi}{4}$
(d) None of these
›Reveal solutionSolution

The principal value of $\tan^{-1}x$ always lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ .

We need $y$ such that $\tan y = -1$ and $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\cos^{-1} x$
(a) $[0, \pi]$
(b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
(c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
(d) None of these
›Reveal solutionSolution

The principal value branch of $\cos^{-1}x$ is $[0,\pi]$ by definition.

The function $\cos:[0,\pi]\to[-1,1]$ is a bijection, so its inverse $\cos^{-1}:[-1,1]\to[0,\pi]$ is defined w …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Principal value of cos⁻¹(1/2) is:$
(a) π/2
(b) π/3
(c) π/4
(d) π/6
›Reveal solutionSolution

The principal value of $\cos^{-1}x$ lies in $[0,\pi]$ , and $\cos\left(\frac{\pi}{3}\right) = \frac12$ .

We need $\theta \in [0,\pi]$ such that $\cos\theta = \dfrac12$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{4}$
(d) $\dfrac{\pi}{2}$
›Reveal solutionSolution

Find the angle in the principal-value range $[0,\pi]$ of $\cos^{-1}$ whose cosine equals $\dfrac{\sqrt3}{2}$ .

We need $\theta\in[0,\pi]$ (the principal-value branch of $\cos^{-1}$ ) such that

$\cos\theta = \frac{\sqrt3}{2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The principal value of cot⁻¹(-1/\sqrt3) is ................. .$
(a) π/3
(b) π/4
(c) 2π/3
(d) 4π/3
›Reveal solutionSolution

The principal range of $\cot^{-1}$ is $(0,\pi)$ ; find the angle in that range with cotangent $-1/\sqrt3$ .

We know $\cot(\pi/3)=1/\sqrt3$ . Since the given value is negative and the principal range of $\cot^{-1}$ is $(0,\pi)$ , the required angle lies in the second quadrant, where cotangent is negative.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The principal value of tan^{-1}(-\sqrt{3}) is:$
(a)
(i) \pi/3
(b)
(ii) -\pi/3
(c)
(iii) \pi/6
(d)
(iv) -\pi/6
›Reveal solutionSolution
$\tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}$ — option (ii).

Concept. The principal value of $\tan^{-1} x$ is the unique angle $\theta$ lying in the open interval $\left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$ such that $\tan\theta = x$ . This is the standard NCERT/CBSE principal-value branch that the UBSE Class-12 syllabus also follows.

Why this branch. Tangent is one-to-one on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , so exactly one angle there gives each real value.

Steps.

We need $\theta$ with $\tan\theta = -\sqrt{3}$ and $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ . …
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CBSE 2025Set X11 markMCQ

\sin^{-1} x

List - I	List - II
A) Domain of $\sin^{-1} x$	i) $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$
B) Range of $\tan^{-1} x$	ii) $[0, \pi]$
C) Range of $\cos^{-1} x$	iii) $[-1, 1]$

(a) A-i, B-ii, C-iii

(b) A-iii, B-ii, C-i

(d) A-iii, B-i, C-ii

›Reveal solution

Matching inverse-trig domains/ranges — correct option is (d).

Recall the standard facts: the domain of $\sin^{-1}x$ is $[-1,1]$ (matches iii), the principal range of $\tan^{-1}x$ is the open interval $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ (matches i), and the principal range …

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Q.State True or False: The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ …

Why this question matters

Step‑by‑step reasoning

Method: Reason about which restrictions permit an inverse

Steps

Common Mistakes

Concept and Intuition

Step-by-step solution

Q.State True or False: The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Why this question matters

Step‑by‑step reasoning

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