What Is Cot Inverse x?
Cot Inverse x is the inverse of the cotangent function: it takes a number $x$ and returns the angle whose cotangent is $x$. In symbols, if $\cot\theta = x$, then $\theta = \cot^{-1}x$, also written $\operatorname{arccot} x$. So $\cot^{-1}x$ answers the question "which angle has this cotangent?"
The cotangent of an angle is the ratio of the adjacent side to the opposite side in a right triangle, or equivalently $\cos\theta / \sin\theta$. Because $\cot\theta$ runs through every real value as $\theta$ sweeps across an interval, the arccotangent accepts every real number as input. For the wider family this belongs to, see inverse trigonometric functions.
One caution sits at the centre of this topic. The superscript $-1$ marks an inverse function, not a reciprocal. So $\cot^{-1}x$ is not $\dfrac{1}{\cot x}$; the reciprocal of cotangent is tangent, a completely different object.
What Are The Domain And Range Of Cot Inverse x?
The domain of Cot Inverse x is all real numbers, from $-\infty$ to $+\infty$. Every real number is the cotangent of some angle, so every real number has an arccotangent.
The range is where honesty matters, because two conventions are in active use and they disagree for negative inputs. The range is called the principal branch: the single slice of angles we agree to return so that the inverse is a function.
Convention A, range $(0, \pi)$. Used by India's NCERT, by Wolfram MathWorld, and by most school and reference sources. Here $\cot^{-1}x$ is a smooth, continuous, decreasing curve, and negative inputs return obtuse angles between $\frac{\pi}{2}$ and $\pi$. This is the convention this article uses unless stated otherwise.
Convention B, range $\left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$. Used by Mathematica and some calculus texts, chosen so that $\cot^{-1}x = \arctan\frac{1}{x}$. Here negative inputs return negative angles, and the graph has a jump at $x = 0$.
The split is not an error in one source or the other; it is a genuine choice about which property to keep. Convention A keeps the graph continuous. Convention B keeps a clean link to $\arctan\frac{1}{x}$. The principal value of trigonometric functions page treats this idea across the whole inverse family, and the general domain and range of trigonometric functions reference sits alongside it.
Table: The two principal-branch conventions for Cot Inverse x, compared.
Feature | Convention A: $(0, \pi)$ | Convention B: $\left(-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]$ |
|---|---|---|
Who uses it | NCERT, Wolfram MathWorld, most textbooks | Mathematica, some US calculus texts |
Continuous? | Yes, one smooth curve | No, jumps at $x=0$ |
$\cot^{-1}(-1)$ | $\frac{3\pi}{4} = 135^\circ$ | $-\frac{\pi}{4} = -45^\circ$ |
Identity it preserves | $\cot^{-1}x = \frac{\pi}{2} - \arctan x$ | $\cot^{-1}x = \arctan\frac{1}{x}$ |
Always check which convention a textbook, exam board, or piece of software assumes before you trust a negative answer.
What Does The Graph Of Cot Inverse x Look Like?
In Convention A the graph of Cot Inverse x is a single decreasing curve. As $x$ runs from $-\infty$ to $+\infty$, the output falls steadily from $\pi$ down towards $0$, passing through $\frac{\pi}{2}$ at $x = 0$. The lines $y = 0$ and $y = \pi$ are horizontal asymptotes that the curve approaches but never reaches.
The function is strictly decreasing everywhere, which is why the derivative (below) is always negative. It is neither even nor odd, and it has no repetition, since the whole point of the principal branch is to pick one output per input.
Where Does Each Value Of Cot Inverse x Sit On The Unit Circle?
Every arccotangent output is an angle you can place on the unit circle. Positive inputs land in the first quadrant, between $0$ and $\frac{\pi}{2}$; the value $x = 0$ lands exactly on the top at $\frac{\pi}{2}$; and negative inputs (in Convention A) land in the second quadrant, between $\frac{\pi}{2}$ and $\pi$, where cosine is negative and sine is positive, making cotangent negative.
What Are The Key Values Of Cot Inverse x?
These are the standard values every student should recognise on sight, shown in both radians and degrees. They use Convention A, range $(0, \pi)$. Each one is exact, drawn from the cotangents of the special angles.
Table: Key values of Cot Inverse x in radians and degrees (range $(0,\pi)$).
$x$ | $\cot^{-1}x$ (radians) | $\cot^{-1}x$ (degrees) |
|---|---|---|
$\sqrt{3}$ | $\frac{\pi}{6}$ | $30^\circ$ |
$1$ | $\frac{\pi}{4}$ | $45^\circ$ |
$\frac{1}{\sqrt{3}}$ | $\frac{\pi}{3}$ | $60^\circ$ |
$0$ | $\frac{\pi}{2}$ | $90^\circ$ |
$-\frac{1}{\sqrt{3}}$ | $\frac{2\pi}{3}$ | $120^\circ$ |
$-1$ | $\frac{3\pi}{4}$ | $135^\circ$ |
$-\sqrt{3}$ | $\frac{5\pi}{6}$ | $150^\circ$ |
Each row is worth checking against the direct value pages: for example $\cot\frac{\pi}{4} = 1$ (see cot pi/4), $\cot\frac{\pi}{6} = \sqrt{3}$ (see cot pi/6), and $\cot\frac{\pi}{2} = 0$ (see cot pi/2). Reading a value forwards and then backwards is the fastest way to feel what an inverse function does.
How Is Cot Inverse x Related To Arctan?
In Convention A there is a clean bridge between arccotangent and arctan:
$$\cot^{-1}x = \frac{\pi}{2} - \arctan x$$
Here is why it holds. Let $y = \cot^{-1}x$, so $\cot y = x$ with $y$ in $(0, \pi)$. Using the cofunction identity $\cot y = \tan\left(\frac{\pi}{2} - y\right)$:
$$\tan\left(\frac{\pi}{2} - y\right) = x$$
Since $y$ lies in $(0, \pi)$, the angle $\frac{\pi}{2} - y$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, which is exactly the principal range of arctan. So we can take arctan of both sides:
$$\frac{\pi}{2} - y = \arctan x$$
$$y = \frac{\pi}{2} - \arctan x$$
That last line is the identity. It rests on the cofunction identities, and it gives a quick way to compute any arccotangent from an arctangent you already know.
How Do You Find The Derivative Of Cot Inverse x?
The derivative of Cot Inverse x is:
$$\frac{d}{dx}\cot^{-1}x = -\frac{1}{1+x^2}$$
It comes straight from implicit differentiation. Let $y = \cot^{-1}x$, so $\cot y = x$. Differentiate both sides with respect to $x$:
$$-\csc^2 y \cdot \frac{dy}{dx} = 1$$
$$\frac{dy}{dx} = -\frac{1}{\csc^2 y}$$
Now use the Pythagorean identity $\csc^2 y = 1 + \cot^2 y$, and recall $\cot y = x$:
$$\frac{dy}{dx} = -\frac{1}{1 + \cot^2 y} = -\frac{1}{1+x^2}$$
The result is negative for every $x$, which matches the graph: Cot Inverse x always decreases. It differs from the derivative of arctan only by a sign, since $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2}$. Compare it with the derivative of tan inverse x and the wider rules in differentiation of trigonometric functions.
How Do You Solve Cot Inverse x Problems?
Work in Convention A, range $(0, \pi)$, unless a question states otherwise. The method is the same each time: read the input, find the special angle whose cotangent matches it, and check the quadrant.
Example 1: Find $\cot^{-1}(\sqrt{3})$.
Ask which angle in $(0, \pi)$ has cotangent $\sqrt{3}$. Since $\cot\frac{\pi}{6} = \sqrt{3}$, the answer sits in the first quadrant.
Final answer: $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6} = 30^\circ$.
Example 2: Find $\cot^{-1}(-1)$.
A negative input means an obtuse angle in the second quadrant. The reference angle has cotangent $1$, which is $\frac{\pi}{4}$, so the answer is $\pi - \frac{\pi}{4}$.
Final answer: $\cot^{-1}(-1) = \frac{3\pi}{4} = 135^\circ$ in Convention A. (Convention B would give $-\frac{\pi}{4}$.)
Example 3: Use the arctan relation to find $\cot^{-1}(1)$.
Apply $\cot^{-1}x = \frac{\pi}{2} - \arctan x$ with $x = 1$. Since $\arctan 1 = \frac{\pi}{4}$:
$$\cot^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$
Final answer: $\cot^{-1}(1) = \frac{\pi}{4} = 45^\circ$.
Why Is Cot Inverse x A Decreasing Function?
The steady fall of the curve is not a coincidence; it follows from how cotangent behaves on the principal branch.
Cotangent decreases across $(0, \pi)$. Starting just above $0$, $\cot\theta$ is very large and positive; by $\frac{\pi}{2}$ it has fallen to $0$; approaching $\pi$ it drops towards $-\infty$. A function that only decreases has an inverse that only decreases.
Large inputs mean small angles. A big positive $x$ means a very small angle near $0$, because only a near-zero angle has a huge cotangent. As $x$ shrinks and turns negative, the angle grows toward $\pi$.
The sign of the derivative confirms it. Since $1 + x^2$ is always positive, $-\frac{1}{1+x^2}$ is always negative, so the slope is never zero and never positive.
The unit-circle picture ties these together: positive inputs sit in the first quadrant near the horizontal axis, and negative inputs swing up into the second quadrant. This is the standard inverse trigonometric ratios behaviour applied to cotangent.
Who Invented The Notation For Cot Inverse x?
The idea of an inverse trigonometric value is old, but the compact symbol $\cot^{-1}$ is surprisingly recent, and its design is the source of the topic's most common confusion.
Leonhard Euler had already worked extensively with the "arc" functions in the 1700s, treating them as arc lengths on the circle, which is why the older name $\operatorname{arccot} x$ is still standard today and sits happily beside Herschel's $\cot^{-1}x$.
Where Is Cot Inverse x Used In The Real World?
Recovering an angle from a ratio is a routine need, and arccotangent is the tool whenever that ratio is naturally written as horizontal-over-vertical.
Engineering and construction: finding the incline of a ramp, road, or roof from its run and rise, where the run-to-rise ratio is exactly a cotangent.
Navigation and surveying: turning a horizontal distance and a height into an angle of elevation or depression for line-of-sight calculations.
Physics: resolving forces and projectile paths, where an angle must be extracted from a ratio of horizontal and vertical components.
Computer graphics: computing view and camera angles from coordinate ratios, often through the closely related arctangent.
Antenna and signal design: setting tilt angles from geometric ratios of distance and height.
Across all of them the pattern is the same: a measurable ratio goes in, and a usable angle comes out.
What Are The Most Common Mistakes With Cot Inverse x?
These three errors account for most lost marks, and each is confirmed by the documented disagreements between textbooks and software.
Reading $\cot^{-1}x$ as $\dfrac{1}{\cot x}$.
Where it slips in:
A student sees the $-1$ exponent and treats it as a reciprocal, the way $\cot^{2}x$ means $(\cot x)^2$.
Don't do this:
Do not write $\cot^{-1}x = \dfrac{1}{\cot x}$. The reciprocal of cotangent is tangent, not the arccotangent.
The correct way:
Read $\cot^{-1}x$ as "the angle whose cotangent is $x$". If you want the reciprocal instead, that is $\tan x$, and the two are unrelated.
Ignoring the range convention on negative inputs.
Where it slips in:
A student computes $\cot^{-1}(-1)$ as $-\frac{\pi}{4}$ from a calculator, then loses marks in an NCERT-style exam that expects $\frac{3\pi}{4}$.
Don't do this:
Do not assume every source uses the same principal branch. Software often uses Convention B while school exams use Convention A.
The correct way:
Fix the convention first. In range $(0, \pi)$, a negative input gives an obtuse angle between $\frac{\pi}{2}$ and $\pi$, so $\cot^{-1}(-1) = \frac{3\pi}{4}$.
Placing a negative input in the wrong quadrant.
Where it slips in:
A student finds the reference angle correctly but then reports the first-quadrant angle for a negative $x$.
Don't do this:
Do not return an acute angle for a negative input in Convention A. That ignores where cotangent is negative.
The correct way:
For negative $x$ in range $(0, \pi)$, subtract the reference angle from $\pi$. For $x = -\sqrt{3}$, the reference angle is $\frac{\pi}{6}$, so the answer is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Practice Problems On Cot Inverse x
Work in range $(0, \pi)$ unless told otherwise. Answers follow each line.
Find $\cot^{-1}(0)$.
(Answer: $\frac{\pi}{2} = 90^\circ$.)Find $\cot^{-1}\left(\frac{1}{\sqrt{3}}\right)$.
(Answer: $\frac{\pi}{3} = 60^\circ$.)Find $\cot^{-1}(-\sqrt{3})$.
(Answer: $\frac{5\pi}{6} = 150^\circ$.)Use $\cot^{-1}x = \frac{\pi}{2} - \arctan x$ to find $\cot^{-1}(0)$.
(Answer: $\frac{\pi}{2} - 0 = \frac{\pi}{2}$.)Find the derivative $\frac{d}{dx}\cot^{-1}x$ at $x = 2$.
(Answer: $-\frac{1}{1+2^2} = -\frac{1}{5}$.)Evaluate $\cot\left(\cot^{-1}(7)\right)$.
(Answer: $7$, since cotangent undoes its own inverse for every real input.)
Where Should You Go Next After Cot Inverse x?
Cot Inverse x is one door into the inverse-function side of trigonometry, and a few natural next steps open from here.
Inverse trigonometric functions. The full family (arcsin, arccos, arctan, and the reciprocals) with their domains and principal ranges side by side.
Sec inverse x. The sibling reciprocal-function inverse, whose range convention carries the same kind of subtlety.
Trigonometric ratios. Go back to the six ratios themselves, so the forward direction (angle to ratio) is solid before you reverse it.
If your child is building these foundations, a live Bhanzu trainer teaches inverse trigonometry starting from the "why" (recovering angles from ratios) in the Bhanzu trigonometry program.
Was this article helpful?
Your feedback helps us write better content