State True or False: The least numerical value, either positive or negative of angle $\theta$ is called principal value of the inverse trigonometric function.

State True or False: The least numerical value, either positive or negative of angle θ is called principal val

NCERT Exemplar · Q52

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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 agrees with the standard definition: the principal value is the numerically smallest angle (positive or negative) whose trig value matches — so it is True.

The idea

A trig equation like $\sin\theta=\frac{1}{2}$ has infinitely many solutions. To make the inverse a genuine function we pick one representative, and the convention is the angle of least numerical magnitude — the one closest to $0$ — which is precisely how the principal-value branches are defined.

Why the statement is correct

Each inverse function's principal range is built around $0$ so that it returns the numerically smallest angle:

$\sin^{-1}$ : range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , e.g. $\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$ and $\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$ .
$\tan^{-1}$ : range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , e.g. $\tan^{-1}(-1)=-\frac{\pi}{4}$ .

In every case the chosen angle is the one of least magnitude, positive or negative according to the sign — exactly the description in the statement.

A note on $\cos^{-1}$ …

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

Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQ
$\tan^{-1}(-\sqrt{3})$
(a) $\frac{\pi}{3}$
(b) $-\frac{\pi}{3}$
(c) $-\frac{\pi}{4}$
(d) $\frac{\pi}{4}$
›Reveal solutionSolution

Use $\tan^{-1}(-x)=-\tan^{-1}(x)$ and the known value $\tan^{-1}\sqrt3=\dfrac{\pi}{3}$ , checking the result lies in the principal-value range.

The principal value branch of $\tan^{-1}$ is $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ , over which $\tan^{-1}$ is an odd function:

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

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 markMCQ
$\cos^{-1}\left(-\dfrac{1}{2}\right)$
(a) $\pi/3$
(b) $2\pi/3$
(c) $\pi/6$
(d) $5\pi/6$
›Reveal solutionSolution

$\cos^{-1}x$ takes values only in $[0,\pi]$ ; find the angle in that range whose cosine is $-\tfrac12$ .

…

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 mark
$\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$
›Reveal solutionSolution

Evaluate each inverse-trig term on its principal branch, then subtract.

$\tan^{-1}\sqrt3$ : principal value in $\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$ with $\tan\theta=\sqrt3$ gives $\theta=\dfrac{\pi}{3}$ .

…

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Q.State True or False: The least numerical value, either positive or negative of angle $\theta$ is called principal value of the inverse trigonometric function.

Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ …

The idea

Why the statement is correct

A note on $\cos^{-1}$ …

Method: Test a definition statement against the standard branches

Steps

Common Mistakes

Q.State True or False: The least numerical value, either positive or negative of angle θ\thetaθ is called principal value of the inverse trigonometric function.

The idea

Why the statement is correct

A note on cos⁡−1\cos^{-1}cos−1 …

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Q.State True or False: The least numerical value, either positive or negative of angle $\theta$ is called principal value of the inverse trigonometric function.

A note on $\cos^{-1}$ …

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