Find the value of tan⁻¹(-dfrac1√(3)) + cot⁻¹(dfrac1√(3)) + tan⁻¹[sin(-(π)/(2))]. OR Find the domain of the fun

Question

\tan^{-1}\left(-\dfrac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\dfrac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\dfrac{\pi}{2}\right)\right]

(OR)

f(x) = \sin^{-1}(x^2 - 4)

CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Domain Of Composite Function

Domain of a Composite Function

Picture a building with two doors: the first opens with a blue pass, the second with a red pass. Walking through $f(g(x))$ means passing the inner door $g$ first, then the outer door $f$ . The domain of the composite is simply: which inputs make it through both doors?

The intuition

Take $f(x) = \sqrt{x}$ and $g(x) = x - 5$ , so $f(g(x)) = \sqrt{x-5}$ .

The inner $g(x) = x - 5$ accepts every real number.
The outer $f$ accepts only non-negative inputs.

So the real question is: which $x$ make $g(x)$ land inside the domain of $f$ ? The domain of the composite is not just the domain of $g$ , nor just the domain of $f$ — it is the overlap seen through $g$ .

The precise statement

\text{Domain}(f \circ g) = \{\, x \in \text{Domain}(g) \;\mid\; g(x) \in \text{Domain}(f) \,\}.

Two steps, in order:

Keep only the $x$ that $g$ can handle.
Among those, keep only the $x$ for which $g(x)$ is something $f$ can handle.

Watch out

A frequent mistake is to restrict $x$ using the domain of $f$ directly. The restriction comes from $g(x)$ lying in $\text{Domain}(f)$ , not from $x$ itself.

A worked check

For $f(x) = \sqrt{x}$ , $g(x) = \dfrac{1}{x-1}$ :

Domain of $g$ : $x \ne 1$ .
Outer condition: $g(x) \ge 0 \Rightarrow \dfrac{1}{x-1} \ge 0 \Rightarrow x - 1 > 0 \Rightarrow x > 1$ .

So $\text{Domain}(f\circ g) = (1, \infty)$ — the condition from $f$ already excludes $x = 1$ .

Order matters …

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Part (a): the sum of the three inverse-trig values is $-\dfrac{\pi}{12}$ . Part (b): the domain of $\sin^{-1}(x^2-4)$ is $[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5]$ and its range is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ .

Part (a)

Concept

Evaluate each inverse function within its principal-value branch: $\tan^{-1}\in(-\tfrac\pi2,\tfrac\pi2)$ , $\cot^{-1}\in(0,\pi)$ .

Steps

$\tan^{-1}\!\Big(-\tfrac{1}{\sqrt3}\Big)=-\dfrac{\pi}{6}$ (since $\tan\tfrac\pi6=\tfrac1{\sqrt3}$ ).
$\cot^{-1}\!\Big(\tfrac{1}{\sqrt3}\Big)=\dfrac{\pi}{3}$ (since $\cot\tfrac\pi3=\tfrac1{\sqrt3}$ , and $\tfrac\pi3\in(0,\pi)$ ).
$\sin(-\tfrac{\pi}{2})=-1$ , so $\tan^{-1}(-1)=-\dfrac{\pi}{4}.$
Add: …

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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 68 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.Find the value of tan⁡−1(−13)+cot⁡−1(13)+tan⁡−1[sin⁡(−π2)]\tan^{-1}\left(-\dfrac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\dfrac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\dfrac{\pi}{2}\right)\right]tan−1(−3​1​)+cot−1(3​1​)+tan−1[sin(−2π​)].

Part (a)

Concept

Steps

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Q.Find the value of $\tan^{-1}\left(-\dfrac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\dfrac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\dfrac{\pi}{2}\right)\right]$ .

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