If $\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0$, then value of $x$ is __________.

If cos(tan⁻¹x+cot⁻¹√3)=0, then value of x is _____.

NCERT Exemplar · Q40

\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0

CBSEShort· 1mImportance★★★★★

82% · 89/108 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

The key idea is to use the identity $\tan^{-1}x + \cot^{-1}\sqrt{3} = \frac{\pi}{2}$ (since $\cos\theta = 0$ implies $\theta = \frac{\pi}{2}$ ), then simplify using $\cot^{-1}\sqrt{3} = \frac{\pi}{6}$ , giving $\tan^{-1}x = \frac{\pi}{3}$ , so $x = \sqrt{3}$ .

The problem asks for $x$ when $\cos(\tan^{-1}x + \cot^{-1}\sqrt{3}) = 0$ . This is a composite trigonometric equation. The core idea: $\cos\theta = 0$ means $\theta$ is an odd multiple of $\frac{\pi}{2}$ , but here the angles are principal values (since $\tan^{-1}$ and $\cot^{-1}$ are defined with specific ranges), so we can directly set the sum equal to $\frac{\pi}{2}$ .

Why? Because $\tan^{-1}x$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ and $\cot^{-1}\sqrt{3}$ lies in $(0, \pi)$ . Their sum could be many things, but for the cosine to be zero, the sum must be $\frac{\pi}{2}$ (or $-\frac{\pi}{2}$ , but that’s impossible here — check ranges). So we solve:

Set the argument equal to $\frac{\pi}{2}$ Since $\cos\theta = 0$ implies $\theta = \frac{\pi}{2} + n\pi$ , but the principal values restrict us. $\cot^{-1}\sqrt{3}$ is a positive acute angle: $\cot^{-1}\sqrt{3} = \frac{\pi}{6}$ (since $\cot\frac{\pi}{6} = \sqrt{3}$ ). And $\tan^{-1}x$ is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ . Their sum can only be $\frac{\pi}{2}$ (not $-\frac{\pi}{2}$ or $\frac{3\pi}{2}$ ). So: $\tan^{-1}x + \frac{\pi}{6} = \frac{\pi}{2}$ …

Start your 14-day free trial to unlock the full solution →

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 2020Set 65/2/11 mark
$y = \sec^{-1} x$
(OR)
$\cos^{-1}\left(-\dfrac{1}{2}\right)$
›Reveal solutionSolution
The principal-value branch of $\sec^{-1}x$ has range $[0,\pi]\setminus\{\pi/2\}$ .

$\cos^{-1}(-\tfrac12)=\tfrac{2\pi}{3}$ .

Part (a)

To invert a periodic function we restrict it to an interval on which it is one-one and onto its range; that restricted interval becomes the range of the inverse. For $\sec x=\dfrac{1}{\cos x}$ the standard choice is $[0,\pi]$ , but $\sec x$ is undefined where $\cos x=0$ , i.e. at $x=\tfrac{\pi}{2}$ , so this point must be removed.

On $[0,\tfrac{\pi}{2})$ , $\sec x$ runs from $1$ to $+\infty$ .

On $(\tfrac{\pi}{2},\pi]$ , $\sec x$ runs from $-\infty$ to $-1$ .

Together these cover all outputs with $|x|\ge1$ one-to-one, which is exactly the domain of $\sec^{-1}$ . …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 $\sec^{-1}\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ .

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

Start your 14-day free trial to see the full solution →
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 $x\in[-1,1]$ .

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $\cos\theta = \dfrac12$ .

…

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ39 — The value of sin⁻¹(sin(3π)/(5)) is _____.→Next questionQ41 — The set of values of sec⁻¹((1)/(2)) is _____.

Q.If $\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0$ , then value of $x$ is __________.

Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ …

Method: Solving an equation wrapped in an outer trig function

Steps

Common Mistakes

Part (a)

Concept and Intuition

Step-by-step solution

Q.If cos⁡(tan⁡−1x+cot⁡−13)=0\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0cos(tan−1x+cot−13​)=0, then value of xxx is __________.

🎓Unlock everything free for 14 days

Q.If $\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0$ , then value of $x$ is __________.

Unlock everything free for 14 days