Chapter 2 Inverse Trigonometric Functions (Concepts)

Introduction

The concept of Inverse Trigonometric Functions (also referred to as Arcfunctions) arises from the need to determine the angle when the value of a trigonometric ratio is known. Unlike algebraic functions, trigonometric functions like $\sin x$, $\cos x$, and $\tan x$ are periodic, meaning they repeat their values after specific intervals. This characteristic makes them many-one functions in their natural domains.

For any function to have an inverse, it must satisfy the condition of being a Bijection (both one-one and onto). To achieve this for trigonometric functions, we mathematically restrict their natural domains to specific intervals where they become strictly monotonic (either increasing or decreasing). Within these restricted boundaries, an inverse can be uniquely defined.

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Basic Concept of Invertibility

Let $f$ be a function defined from domain $D_f$ to range $R_f$. If $f$ is bijective, there exists a unique inverse function $f^{-1}$ that maps the range back to the domain. The fundamental relationship is expressed as:

In this transition, the domain of $f$ becomes the range of $f^{-1}$, and the range of $f$ becomes the domain of $f^{-1}$. This interchange of roles is the core principle behind inverse operations.

Graphical Reflection and Symmetry

The relationship between a function and its inverse is beautifully represented through geometry. The graph of $f^{-1}$ is essentially the mirror image of the graph of $f$ with respect to the identity line $y = x$.

Geometric Interpretation:

1. If a point $(a, b)$ satisfies the equation $y = f(x)$, then the point $(b, a)$ must satisfy $x = f^{-1}(y)$.

2. Mathematically, the line $y = x$ acts as a perpendicular bisector for the segment connecting $(a, b)$ and $(b, a)$.

3. This reflection effectively swaps the horizontal and vertical axes, which is why we notice that the $x$-axis labels of a trigonometric graph (angles in radians) appear on the $y$-axis of an inverse trigonometric graph.

Conditions for Existence

To ensure that we only get one output for every input (maintaining the definition of a function), we choose a specific branch for the inverse function. This chosen interval is called the Principal Value Branch.

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Table of Domains and Principal Value Branches

The Domain of an inverse trigonometric function corresponds to the range of the original trigonometric function. However, the Range of an inverse trigonometric function must be restricted to a specific interval to ensure the function remains well-defined and unique. This restricted range is mathematically referred to as the Principal Value Branch.

If $y = f^{-1}(x)$, then the value of $y$ lying in the principal value branch is called the Principal Value of the function. While there are infinitely many intervals where these functions are bijective, the mathematical community has standardized these specific branches for consistency in calculus and engineering.

Summary Table of Domains and Ranges

The following table provides a comprehensive overview of the defined domains and the corresponding principal value branches (ranges) for all six inverse trigonometric functions.

Inverse Function	Domain (Values of $x$)	Principal Value Range (Values of $y$)
$y = \sin^{-1} x$	$[-1, 1]$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$
$y = \cos^{-1} x$	$[-1, 1]$	$[0, \pi]$
$y = \tan^{-1} x$	$\mathbb{R}$	$\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$
$y = \cot^{-1} x$	$\mathbb{R}$	$(0, \pi)$
$y = \sec^{-1} x$	$\mathbb{R} - (-1, 1)$ or $(-\infty, -1] \cup [1, \infty)$	$[0, \pi] - \{ \frac{\pi}{2} \}$
$y = \text{cosec}^{-1} x$	$\mathbb{R} - (-1, 1)$ or $(-\infty, -1] \cup [1, \infty)$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$

Comparative Table of Domains and Ranges

The following table provides a side-by-side comparison. Note that for the original trigonometric functions, we use the Restricted Domain (Principal Value Branch) to ensure they are bijective (one-one and onto), which is a prerequisite for invertibility.

Trigonometric Function	Restricted Domain	Range	Inverse Function	Domain of Inverse	Range (Principal Branch)
$\sin x$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$	$[-1, 1]$	$\sin^{-1} x$	$[-1, 1]$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$
$\cos x$	$[0, \pi]$	$[-1, 1]$	$\cos^{-1} x$	$[-1, 1]$	$[0, \pi]$
$\tan x$	$\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$	$\mathbb{R}$	$\tan^{-1} x$	$\mathbb{R}$	$\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$
$\cot x$	$(0, \pi)$	$\mathbb{R}$	$\cot^{-1} x$	$\mathbb{R}$	$(0, \pi)$
$\sec x$	$[0, \pi] - \{ \frac{\pi}{2} \}$	$\mathbb{R} - (-1, 1)$	$\sec^{-1} x$	$(-\infty, -1] \cup [1, \infty)$	$[0, \pi] - \{ \frac{\pi}{2} \}$
$\text{cosec } x$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$	$\mathbb{R} - (-1, 1)$	$\text{cosec}^{-1} x$	$(-\infty, -1] \cup [1, \infty)$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$

Trigonometric functions are periodic, meaning they repeat their values after a fixed interval (usually $2\pi$ or $\pi$). For example, $\sin(0) = \sin(\pi) = \sin(2\pi) = 0$. This many-one nature prevents the existence of a unique inverse. To define an inverse function, we must restrict the domain to an interval where the function is monotonic (strictly increasing or strictly decreasing), thereby making it one-one and onto (bijective).

1. Logic for Sine and Cosine Restrictions

(a) Restriction for $y = \sin^{-1} x$

The sine function $f(x) = \sin x$ is strictly increasing in the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$. In this specific branch, the function takes all values from $-1$ to $1$ exactly once. While other intervals like $\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$ also satisfy this, $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ is chosen as the Principal Value Branch because it is centered around the origin.

(b) Restriction for $y = \cos^{-1} x$

The choice of $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ fails for the cosine function. This is because cosine is an even function, and it repeats values on either side of the y-axis.

For instance, $\cos(-\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$. Therefore, the function is not one-one in the interval used for sine. To find a branch where cosine is one-one, we select the first quadrant and the second quadrant, i.e., $[0, \pi]$. In this interval, cosine is strictly decreasing from $1$ to $-1$.

2. Exclusions in Reciprocal Inverse Functions

The inverse functions of $\text{cosec } x$ and $\sec x$ require additional restrictions because the original functions are reciprocals that involve division by zero at certain points.

(a) Range of $y = \text{cosec}^{-1} x$

The cosecant function is the reciprocal of sine. In its restricted domain $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, there exists a point $x = 0$ where the sine function vanishes.

Since $\text{cosec } x$ is not defined at $0$, the angle $0$ cannot be included in the output of its inverse. Thus, the range of $\text{cosec}^{-1} x$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$.

(b) Range of $y = \sec^{-1} x$

Similarly, the secant function is the reciprocal of cosine. In the restricted domain $[0, \pi]$, there is a point $x = \frac{\pi}{2}$ where the cosine function is zero.

Because $\sec x$ is undefined at $\frac{\pi}{2}$, this value must be excluded from the range of $\sec^{-1} x$. Therefore, its principal value branch is $[0, \pi] - \{ \frac{\pi}{2} \}$.

3. Domain "Gap" for Secant and Cosecant

Unlike $\sin^{-1} x$ and $\cos^{-1} x$, which accept values between $-1$ and $1$, the functions $\sec^{-1} x$ and $\text{cosec}^{-1} x$ have a domain where values between $-1$ and $1$ are excluded. This is because $|\sin x| \leq 1$ and $|\cos x| \leq 1$, implying their reciprocals must always be greater than or equal to $1$ or less than or equal to $-1$.

Answer:

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Answer:

Inverse Trigonometric Functions

To define the inverse of trigonometric functions, we must restrict their domains to intervals where they are bijective (both one-one and onto). Below is a detailed analysis of all six inverse trigonometric functions, including their restricted domains, principal value branches, and mathematical definitions.

1. Inverse Sine Function ($\sin^{-1} x$)

The Inverse Sine Function, denoted as $\sin^{-1} x$ or $\text{arcsin } x$, is the inverse of the sine function. In its natural domain $(-\infty, \infty)$, the sine function is periodic and thus many-one. For instance, $\sin 0 = \sin \pi = 0$. To define a unique inverse, we must restrict the domain of the sine function to an interval where it is strictly monotonic (strictly increasing or decreasing).

The standard interval chosen is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, known as the Principal Value Branch. In this interval, the sine function is strictly increasing from $-1$ to $1$, making it a bijection.

Mathematical Definition

If we restrict the sine function such that $f: \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \to [-1, 1]$, then its inverse function $\sin^{-1} : [-1, 1] \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ is defined as:

Where the domain and range are specified as follows:

$\bullet$ Domain: $x \in [-1, 1]$

$\bullet$ Range (Principal Value): $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$

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The natural sine function, $f(x) = \sin x$, is defined for all real numbers ($x \in \mathbb{R}$). While it is a perfectly valid function, its periodic nature poses a significant challenge when we attempt to find its inverse. To understand why we must restrict its domain, we must examine its behavior across the entire number line.

1. The Complete Sine Graph

The sine function is a wave that repeats its values every $2\pi$ radians. Below is the graphical representation of the function $y = \sin x$ over a larger interval. Notice how the wave consistently reaches the same heights ($1$) and depths ($-1$) at regular intervals.

2. Failure of the Horizontal Line Test (HLT)

A function is one-one (injective) if and only if every horizontal line intersects its graph at most once. If a horizontal line intersects the graph more than once, it means multiple $x$-values produce the same $y$-value, making it a many-one function.

Observation of the many-one nature:

If we draw a horizontal line at $y = 0.5$, we can observe that it hits the sine curve at infinitely many points:

Because $\frac{\pi}{6} \neq \frac{5\pi}{6}$ but they yield the same output, the function $f(x) = \sin x$ is not one-one over $\mathbb{R}$. In the context of inverses, if we asked "What is the angle whose sine is $0.5$?", there would be no unique answer, which violates the definition of a function.

3. The Solution: Domain Restriction

To ensure a unique inverse exists, we must "cut" a piece of the sine graph that passes the Horizontal Line Test. We look for the largest interval containing the origin where the function is strictly increasing or decreasing and covers the entire range from $-1$ to $1$.

In this restricted interval, as shown in the image below, any horizontal line between $y = -1$ and $y = 1$ will intersect the curve exactly once. This satisfies the condition for the function to be a bijection (one-one and onto), allowing us to define the inverse function $\sin^{-1} x$ uniquely.

4. The Inverse Sine Graph ($y = \sin^{-1} x$)

To plot $y = \sin^{-1} x$, we take the restricted sine graph and swap the axes. In the original sine function, the $x$-axis represents angles (radians) and the $y$-axis represents numerical ratios. In the inverse sine function:

$\bullet$ The $x$-axis now represents the numerical values in the interval $[-1, 1]$.

$\bullet$ The $y$-axis now represents the angles in the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

This results in a curve that is strictly increasing. It starts at the point $(-1, -\frac{\pi}{2})$, passes through the origin $(0, 0)$, and terminates at $(1, \frac{\pi}{2})$.

5. Combined View and Reflectional Symmetry

The most profound property of inverse functions is their Symmetry across the line $y = x$. The graph of $y = \sin^{-1} x$ is the exact mirror image of the restricted graph of $y = \sin x$ reflected through this identity line.

The Mapping of Points

If a point $(a, b)$ lies on the graph of the sine function, it means $\sin(a) = b$. By the definition of an inverse, $\sin^{-1}(b) = a$, which means the point $(b, a)$ must lie on the inverse sine graph. We can observe this mapping with standard values:

Point on $\sin x$ ($a, b$)	Point on $\sin^{-1} x$ ($b, a$)
$(0, 0)$	$(0, 0)$
$\left( \frac{\pi}{6}, \frac{1}{2} \right)$	$\left( \frac{1}{2}, \frac{\pi}{6} \right)$
$\left( \frac{\pi}{2}, 1 \right)$	$\left( 1, \frac{\pi}{2} \right)$
$\left( -\frac{\pi}{2}, -1 \right)$	$\left( -1, -\frac{\pi}{2} \right)$

6. Mathematical Proof of Symmetry

The line $y = x$ serves as the perpendicular bisector for the line segment joining any pair of points $(a, b)$ and $(b, a)$. To verify this, consider the identity:

Applying this to the sine function:

This identity confirms that the composition of the function and its inverse always maps a value back to itself, visually representing the "return trip" across the mirror line $y = x$.

Important Properties of the Inverse Sine Function

The inverse sine function, $y = \sin^{-1} x$, exhibits several unique algebraic and analytical properties. Understanding these properties is vital for simplifying complex trigonometric expressions and solving equations in calculus. These properties are derived from the foundational definitions of the restricted sine function and the principles of inverse mappings.

I. Strictly Increasing Nature (Monotonicity)

The function $y = \sin^{-1} x$ is a strictly increasing function throughout its entire domain $[-1, 1]$. In mathematical terms, this means that for any two values $x_1$ and $x_2$ in the domain:

II. Odd Function Property

The inverse sine function is an odd function. This means its graph is symmetric with respect to the origin. Mathematically, the property is expressed as:

Derivation:

Let $\sin^{-1}(-x) = \theta$. According to the definition of inverse sine:

Multiplying both sides by $-1$:

$x = -\sin \theta$

Using the trigonometric identity $\sin(-\theta) = -\sin \theta$:

$x = \sin(-\theta)$

Now, taking the inverse sine of both sides:

$\sin^{-1} x = -\theta$

Substituting the value of $\theta$ back into the equation:

Rearranging this gives us the final property:

III. Self-Cancellation (Composition Identities)

When a trigonometric function and its inverse are composed, they effectively "cancel" each other out, returning the original value. However, this is strictly governed by the Domain and Range constraints.

1. Case of $\sin(\sin^{-1} x)$

In this composition, the inverse function is applied first. For $\sin^{-1} x$ to be defined, $x$ must be a ratio between $-1$ and $1$.

2. Case of $\sin^{-1}(\sin x)$

In this composition, the sine function is applied first. The final result is $x$ only if $x$ already lies within the principal value branch of the inverse sine function.

If $x$ lies outside this interval, we must first use trigonometric identities to bring the angle into the principal branch before "canceling" the functions.

Answer:

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2. Inverse Cosine Function ($\cos^{-1} x$)

The Inverse Cosine Function, denoted as $\cos^{-1} x$ or $\text{arccos } x$, is the inverse of the cosine function. Like the sine function, the natural cosine function $f(x) = \cos x$ defined over $\mathbb{R}$ is periodic and many-one. For example, $\cos(0) = 1$ and $\cos(2\pi) = 1$. To define its inverse, we must restrict its domain to an interval where the function is strictly monotonic (strictly decreasing).

The standard interval chosen for the cosine function is $[0, \pi]$, which is referred to as the Principal Value Branch. In this interval, the cosine function decreases from $1$ to $-1$, making it a bijective function (one-one and onto).

Mathematical Definition

If we restrict the cosine function such that $f: [0, \pi] \to [-1, 1]$, then its inverse function $\cos^{-1} : [-1, 1] \to [0, \pi]$ is defined as:

Where the domain and range are specified as follows:

$\bullet$ Domain: $x \in [-1, 1]$

$\bullet$ Range (Principal Value): $y \in [0, \pi]$

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The natural cosine function, $f(x) = \cos x$, is defined for all real numbers ($x \in \mathbb{R}$). However, its periodic nature makes it many-one over its entire domain, preventing a unique inverse. We must restrict the domain to ensure that for every value in the range, there is exactly one corresponding angle.

1. The Complete Cosine Graph

The cosine function is a wave that repeats every $2\pi$ radians, starting at its maximum value of $1$ at $x = 0$. Below is the graphical representation of $y = \cos x$. Note how the function oscillates between $1$ and $-1$.

2. Failure of the Horizontal Line Test (HLT)

Since the cosine function is many-one, it fails the Horizontal Line Test. If we draw a horizontal line at $y = 0.5$, it intersects the graph at multiple points:

Because multiple inputs yield the same output, a unique inverse cannot be defined without restricting the domain.

3. The Solution: Domain Restriction

To define $\cos^{-1} x$, we select the interval $[0, \pi]$. In this branch, the function is strictly decreasing. Every horizontal line between $y = -1$ and $y = 1$ intersects this restricted portion exactly once.

4. The Inverse Cosine Graph ($y = \cos^{-1} x$)

The graph of $y = \cos^{-1} x$ is obtained by interchanging the $x$ and $y$ coordinates of the restricted cosine function. In this graph:

$\bullet$ The $x$-axis represents numerical values in $[-1, 1]$.

$\bullet$ The $y$-axis represents angles in $[0, \pi]$.

The curve starts at $(1, 0)$, passes through $(0, \frac{\pi}{2})$, and ends at $(-1, \pi)$. It is a strictly decreasing function.

5. Combined View and Reflectional Symmetry

The graph of $y = \cos^{-1} x$ is the mirror image of the restricted $y = \cos x$ curve about the line $y = x$. If $(a, b)$ is on the cosine curve, then $(b, a)$ is on the inverse cosine curve.

Point on $\cos x$ ($a, b$)	Point on $\cos^{-1} x$ ($b, a$)
$(0, 1)$	$(1, 0)$
$\left( \frac{\pi}{3}, \frac{1}{2} \right)$	$\left( \frac{1}{2}, \frac{\pi}{3} \right)$
$\left( \frac{\pi}{2}, 0 \right)$	$\left( 0, \frac{\pi}{2} \right)$
$(\pi, -1)$	$(-1, \pi)$

6. Mathematical Proof of Symmetry

The reflectional symmetry is confirmed by the composition identity. Since $f^{-1}(f(x)) = x$ for $x \in [0, \pi]$:

This confirms that the two functions are reflections of each other across the identity line $y = x$.

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Important Properties of the Inverse Cosine Function

The inverse cosine function has unique properties that differ from the inverse sine function, particularly regarding its symmetry and monotonicity.

I. Strictly Decreasing Nature

Unlike $\sin^{-1} x$, the function $y = \cos^{-1} x$ is strictly decreasing on $[-1, 1]$. As the input value $x$ increases, the angle $y$ decreases.

II. Negative Argument Property

The inverse cosine function is neither even nor odd. Because its range is $[0, \pi]$, the inverse of a negative value is found in the second quadrant.

Derivation:

Let $\cos^{-1}(-x) = \theta$. By definition:

Multiplying by $-1$:

$x = -\cos \theta$

Using the identity $\cos(\pi - \theta) = -\cos \theta$:

$x = \cos(\pi - \theta)$

Since $\theta \in [0, \pi]$, then $(\pi - \theta)$ also lies in $[0, \pi]$. Taking $\cos^{-1}$ of both sides:

$\cos^{-1} x = \pi - \theta$

Substituting $\theta = \cos^{-1}(-x)$:

III. Self-Cancellation (Composition Identities)

1. Case of $\cos(\cos^{-1} x)$

2. Case of $\cos^{-1}(\cos x)$

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Answer:

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3. Inverse Tangent Function ($\tan^{-1} x$)

The Inverse Tangent Function, denoted as $\tan^{-1} x$ or $\text{arctan } x$, is the inverse of the tangent function. The natural tangent function $f(x) = \tan x$ is periodic with a period of $\pi$. It is defined for all real numbers except odd multiples of $\frac{\pi}{2}$, where it becomes undefined (vertical asymptotes). Because it repeats its values, it is a many-one function. To define its inverse, we must restrict its domain to an interval where it is strictly monotonic (strictly increasing).

The standard interval chosen for the tangent function is $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, which is referred to as the Principal Value Branch. Note that the interval is open because the function is undefined at the endpoints. In this interval, the tangent function increases from $-\infty$ to $\infty$, making it a bijective function.

Mathematical Definition

If we restrict the tangent function such that $f: \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \to \mathbb{R}$, then its inverse function $\tan^{-1} : \mathbb{R} \to \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ is defined as:

Where the domain and range are specified as follows:

$\bullet$ Domain: $x \in \mathbb{R}$ (All real numbers)

$\bullet$ Range (Principal Value): $y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$

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The natural tangent function, $f(x) = \tan x$, is defined for $x \in \mathbb{R} - \{ (2n+1)\frac{\pi}{2} : n \in \mathbb{Z} \}$. Its periodic nature across multiple branches makes it many-one. For a unique inverse to exist, we must focus on a single branch that covers the entire range of real numbers.

1. The Complete Tangent Graph

The tangent graph consists of infinitely many identical branches separated by vertical asymptotes. Below is the graphical representation of $y = \tan x$. Observe how the function repeats itself every $\pi$ radians.

2. Failure of the Horizontal Line Test (HLT)

The tangent function fails the Horizontal Line Test because any horizontal line $y = k$ will intersect the graph at infinitely many points due to its periodicity. For example, if $y = 1$:

Since multiple inputs result in the same output, the function is many-one over its natural domain.

3. The Solution: Domain Restriction

To define $\tan^{-1} x$, we restrict the domain to the principal branch $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$. In this branch, the function is strictly increasing and passes the Horizontal Line Test perfectly.

4. The Inverse Tangent Graph ($y = \tan^{-1} x$)

The graph of $y = \tan^{-1} x$ is created by swapping the $x$ and $y$ axes. A unique feature of this graph is that the vertical asymptotes of the tangent function become horizontal asymptotes for the inverse function.

$\bullet$ The $x$-axis extends from $-\infty$ to $+\infty$.

$\bullet$ The $y$-axis is bounded between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

The curve passes through the origin $(0, 0)$ and approaches the lines $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ as $x \to \infty$ and $x \to -\infty$ respectively, but never touches them.

5. Combined View and Reflectional Symmetry

The inverse tangent curve is the reflection of the restricted tangent curve across the line $y = x$. We can see the coordinate swap in standard values:

Point on $\tan x$ ($a, b$)	Point on $\tan^{-1} x$ ($b, a$)
$(0, 0)$	$(0, 0)$
$\left( \frac{\pi}{4}, 1 \right)$	$\left( 1, \frac{\pi}{4} \right)$
$\left( \frac{\pi}{3}, \sqrt{3} \right)$	$\left( \sqrt{3}, \frac{\pi}{3} \right)$
$\left( -\frac{\pi}{4}, -1 \right)$	$\left( -1, -\frac{\pi}{4} \right)$

6. Mathematical Proof of Symmetry

Reflective symmetry is verified by the fact that if $y = \tan^{-1} x$, then $\tan y = x$. This swap of variables represents a flip over the identity line:

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Important Properties of the Inverse Tangent Function

The properties of $\tan^{-1} x$ make it one of the most useful inverse trigonometric functions in calculus, especially for integration and describing physical slopes.

I. Strictly Increasing Nature

The function $y = \tan^{-1} x$ is strictly increasing on its entire domain $(-\infty, \infty)$. As $x$ increases from negative infinity to positive infinity, the angle $y$ increases from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

II. Odd Function Property

The inverse tangent function is an odd function, meaning it is symmetric about the origin.

Derivation:

Let $\tan^{-1}(-x) = \theta$. By definition:

Multiplying by $-1$:

$x = -\tan \theta$

Using the identity $\tan(-\theta) = -\tan \theta$:

$x = \tan(-\theta)$

Taking $\tan^{-1}$ of both sides:

$\tan^{-1} x = -\theta$

Substituting $\theta = \tan^{-1}(-x)$ back:

III. Self-Cancellation (Composition Identities)

1. Case of $\tan(\tan^{-1} x)$

2. Case of $\tan^{-1}(\tan x)$

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Answer:

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4. Inverse Cotangent Function ($\cot^{-1} x$)

The Inverse Cotangent Function, denoted as $\cot^{-1} x$ or $\text{arccot } x$, is the inverse of the cotangent function. The natural cotangent function $f(x) = \cot x$ is periodic with a period of $\pi$. It is defined for all real numbers except multiples of $\pi$ ($n\pi$), where the sine function is zero, making the cotangent ratio undefined. Since it repeats its values in every interval of $\pi$, it is a many-one function. To define a unique inverse, we restrict its domain to a single branch where it is strictly monotonic (strictly decreasing).

The standard interval chosen for the cotangent function is $(0, \pi)$, known as the Principal Value Branch. This interval is open because the function is undefined at $0$ and $\pi$. Within this range, the cotangent function decreases from $\infty$ to $-\infty$, covering all real numbers exactly once.

Mathematical Definition

If we restrict the cotangent function such that $f: (0, \pi) \to \mathbb{R}$, then its inverse function $\cot^{-1} : \mathbb{R} \to (0, \pi)$ is defined as:

Where the domain and range are specified as follows:

$\bullet$ Domain: $x \in \mathbb{R}$ (All real numbers)

$\bullet$ Range (Principal Value): $y \in (0, \pi)$

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The natural cotangent function, $f(x) = \cot x$, is valid for $x \in \mathbb{R} - \{ n\pi : n \in \mathbb{Z} \}$. Like other trigonometric functions, its periodic nature across the number line prevents it from having a global inverse.

1. The Complete Cotangent Graph

The graph of $y = \cot x$ consists of multiple disjoint branches separated by vertical asymptotes at $x = \dots, -\pi, 0, \pi, 2\pi, \dots$. Every branch is an identical copy of the other, shifted by $\pi$.

2. Failure of the Horizontal Line Test (HLT)

The cotangent function fails the Horizontal Line Test because it is many-one. A horizontal line drawn at $y = 0$ (the x-axis) will intersect the graph at $\frac{\pi}{2}, \frac{3\pi}{2}, -\frac{\pi}{2}$, etc.

Because multiple inputs produce the same output, the function is not invertible over its entire domain.

3. The Solution: Domain Restriction

To define $\cot^{-1} x$, we restrict the domain to the interval $(0, \pi)$. In this specific branch, the function is strictly decreasing and every $y$-value corresponds to exactly one $x$-value.

4. The Inverse Cotangent Graph ($y = \cot^{-1} x$)

The graph of $y = \cot^{-1} x$ is obtained by reflecting the restricted cotangent curve. In this function, the vertical asymptotes of the original function ($x=0$ and $x=\pi$) transform into horizontal asymptotes ($y=0$ and $y=\pi$).

$\bullet$ The $x$-axis spans from $-\infty$ to $+\infty$.

$\bullet$ The $y$-axis is bounded between $0$ and $\pi$.

The curve starts from a high value near $y = \pi$ as $x \to -\infty$, passes through $(0, \frac{\pi}{2})$, and approaches $y = 0$ as $x \to \infty$.

5. Combined View and Reflectional Symmetry

The graph of $y = \cot^{-1} x$ is a reflection of the restricted $y = \cot x$ across the identity line $y = x$. The coordinate interchange is visible in the following values:

Point on $\cot x$ ($a, b$)	Point on $\cot^{-1} x$ ($b, a$)
$\left( \frac{\pi}{4}, 1 \right)$	$\left( 1, \frac{\pi}{4} \right)$
$\left( \frac{\pi}{2}, 0 \right)$	$\left( 0, \frac{\pi}{2} \right)$
$\left( \frac{3\pi}{4}, -1 \right)$	$\left( -1, \frac{3\pi}{4} \right)$

6. Mathematical Proof of Symmetry

Since the inverse function effectively swaps the input and output, the symmetry is defined by the identity:

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Important Properties of the Inverse Cotangent Function

The inverse cotangent function shares properties with $\cos^{-1} x$ regarding its range and behavior with negative arguments.

I. Strictly Decreasing Nature

The function $y = \cot^{-1} x$ is strictly decreasing on its entire domain $\mathbb{R}$. As the input $x$ increases from negative to positive infinity, the output angle $y$ decreases from $\pi$ towards $0$.

II. Negative Argument Property

Like $\cos^{-1} x$, the inverse cotangent function is neither even nor odd. The inverse of a negative value is mapped to the second quadrant.

Derivation:

Let $\cot^{-1}(-x) = \theta$. This implies $\cot \theta = -x$, where $\theta \in (0, \pi)$.

$x = -\cot \theta$

Using the identity $\cot(\pi - \theta) = -\cot \theta$:

$x = \cot(\pi - \theta)$

Since $\theta \in (0, \pi)$, then $(\pi - \theta)$ also lies in $(0, \pi)$. Taking the inverse cotangent:

$\cot^{-1} x = \pi - \theta$

III. Self-Cancellation (Composition Identities)

1. Case of $\cot(\cot^{-1} x)$

2. Case of $\cot^{-1}(\cot x)$

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Answer:

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5. Inverse Secant Function ($\sec^{-1} x$)

The Inverse Secant Function, denoted as $\sec^{-1} x$ or $\text{arcsec } x$, is the inverse of the secant function. The natural secant function $f(x) = \sec x$ is defined as the reciprocal of the cosine function ($1/\cos x$). It is periodic with a period of $2\pi$. Since it repeats its values, it is a many-one function. To define its inverse, we must restrict its domain to an interval where the function is strictly monotonic.

The standard interval chosen for the secant function is $[0, \pi]$ excluding the point where the function is undefined, which is $\frac{\pi}{2}$ (since $\cos \frac{\pi}{2} = 0$). This interval $[0, \pi] - \{ \frac{\pi}{2} \}$ is referred to as the Principal Value Branch. In this restricted domain, the secant function is bijective, allowing a unique inverse to exist.

Mathematical Definition

If we restrict the secant function such that $f: [0, \pi] - \{ \frac{\pi}{2} \} \to \mathbb{R} - (-1, 1)$, then its inverse function $\sec^{-1} : \mathbb{R} - (-1, 1) \to [0, \pi] - \{ \frac{\pi}{2} \}$ is defined as:

Where the domain and range are specified as follows:

$\bullet$ Domain: $x \in (-\infty, -1] \cup [1, \infty)$ or $|x| \geq 1$

$\bullet$ Range (Principal Value): $y \in [0, \pi] - \{ \frac{\pi}{2} \}$

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The natural secant function, $f(x) = \sec x$, is defined for all real numbers except odd multiples of $\frac{\pi}{2}$. Its periodic nature across the coordinate plane makes it many-one over its entire domain. For a unique inverse to be valid, we must restrict the domain to a specific branch that passes the Horizontal Line Test.

1. The Complete Secant Graph

The graph of $y = \sec x$ consists of multiple U-shaped and inverted U-shaped branches separated by vertical asymptotes. Below is the graphical representation. Note how the function never takes values between $-1$ and $1$.

2. Failure of the Horizontal Line Test (HLT)

The secant function fails the Horizontal Line Test because it is many-one. A horizontal line drawn at $y = 2$ will intersect the graph at $x = \frac{\pi}{3}, \frac{5\pi}{3}, -\frac{\pi}{3}$, etc.

Because multiple inputs yield the same output, the function is not invertible in its natural state.

3. The Solution: Domain Restriction

To define $\sec^{-1} x$, we restrict the domain to $[0, \pi] - \{ \frac{\pi}{2} \}$. In this restricted branch, the function is strictly increasing on both sub-intervals and satisfies the condition of being one-one.

4. The Inverse Secant Graph ($y = \sec^{-1} x$)

The graph of $y = \sec^{-1} x$ is obtained by interchanging the $x$ and $y$ axes of the restricted secant function. The vertical asymptote at $x = \frac{\pi}{2}$ for $\sec x$ becomes a horizontal asymptote at $y = \frac{\pi}{2}$ for $\sec^{-1} x$.

$\bullet$ The $x$-axis contains values from $(-\infty, -1]$ and $[1, \infty)$.

$\bullet$ The $y$-axis contains angles in $[0, \pi]$ excluding $\frac{\pi}{2}$.

5. Combined View and Reflectional Symmetry

The graph of $y = \sec^{-1} x$ is the mirror image of the restricted $y = \sec x$ curve across the line $y = x$. The coordinate interchange is shown below:

Point on $\sec x$ ($a, b$)	Point on $\sec^{-1} x$ ($b, a$)
$(0, 1)$	$(1, 0)$
$\left( \frac{\pi}{3}, 2 \right)$	$\left( 2, \frac{\pi}{3} \right)$
$\left( \frac{2\pi}{3}, -2 \right)$	$\left( -2, \frac{2\pi}{3} \right)$
$(\pi, -1)$	$(-1, \pi)$

6. Mathematical Proof of Symmetry

As per the property of inverse functions, if $y = \sec^{-1} x$, then $\sec y = x$. This swap represents a reflection over the line $y = x$. This is verified by the identity:

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Important Properties of the Inverse Secant Function

The properties of the inverse secant function are closely linked to those of the inverse cosine function due to their reciprocal relationship.

I. Strictly Increasing Nature

The function $y = \sec^{-1} x$ is strictly increasing on both parts of its domain ($(-\infty, -1]$ and $[1, \infty)$). As $x$ increases, the angle $y$ also increases within its respective range segments.

II. Negative Argument Property

The inverse secant function follows the same second-quadrant rule as $\cos^{-1} x$ and $\cot^{-1} x$ for negative inputs.

Derivation:

Let $\sec^{-1}(-x) = \theta$. This implies $\sec \theta = -x$, where $\theta \in [0, \pi] - \{ \frac{\pi}{2} \}$.

$x = -\sec \theta$

Using the identity $\sec(\pi - \theta) = -\sec \theta$:

$x = \sec(\pi - \theta)$

Taking the inverse secant of both sides:

$\sec^{-1} x = \pi - \theta$

Substituting $\theta = \sec^{-1}(-x)$:

III. Self-Cancellation (Composition Identities)

1. Case of $\sec(\sec^{-1} x)$

2. Case of $\sec^{-1}(\sec x)$

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Answer:

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6. Inverse Cosecant Function ($\text{cosec}^{-1} x$)

The Inverse Cosecant Function, denoted as $\text{cosec}^{-1} x$ or $\text{arccosec } x$, is the inverse of the cosecant function. The natural cosecant function $f(x) = \text{cosec } x$ is defined as the reciprocal of the sine function ($1/\sin x$). It is periodic with a period of $2\pi$. Since it repeats its values periodically, it is a many-one function. To define its inverse, we must restrict its domain to an interval where the function is strictly monotonic (strictly decreasing).

The standard interval chosen for the cosecant function is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ excluding the point where the function is undefined, which is $0$ (since $\sin 0 = 0$). This interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$ is referred to as the Principal Value Branch. In this restricted domain, the cosecant function becomes a bijection, allowing a unique inverse to exist.

Mathematical Definition

If we restrict the cosecant function such that $f: \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \to \mathbb{R} - (-1, 1)$, then its inverse function $\text{cosec}^{-1} : \mathbb{R} - (-1, 1) \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$ is defined as:

Where the domain and range are specified as follows:

$\bullet$ Domain: $x \in (-\infty, -1] \cup [1, \infty)$ or $|x| \geq 1$

$\bullet$ Range (Principal Value): $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$

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The natural cosecant function, $f(x) = \text{cosec } x$, is defined for all real numbers except multiples of $\pi$ ($n\pi$). Due to its periodic nature, it takes the same value at infinitely many points. To define a valid inverse function, we must apply the Horizontal Line Test and restrict the function to a branch where it is one-one and onto.

1. The Complete Cosecant Graph

The graph of $y = \text{cosec } x$ consists of disjoint parabolic-like branches separated by vertical asymptotes at $x = 0, \pm\pi, \pm 2\pi, \dots$. Observe that the function never takes any value in the open interval $(-1, 1)$.

2. Failure of the Horizontal Line Test (HLT)

The cosecant function fails the Horizontal Line Test because it is many-one. For example, a horizontal line drawn at $y = 2$ intersects the graph at $x = \frac{\pi}{6}, \frac{5\pi}{6}, \dots$.

Since different inputs yield the same output, the global inverse does not exist.

3. The Solution: Domain Restriction

By restricting the domain to $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$, we obtain a strictly decreasing branch. In this interval, every $y$-value corresponds to exactly one $x$-value, making the function invertible.

4. The Inverse Cosecant Graph ($y = \text{cosec}^{-1} x$)

The graph of $y = \text{cosec}^{-1} x$ is the reflection of the restricted cosecant function across the identity line $y = x$. The vertical asymptote at $x = 0$ for $\text{cosec } x$ becomes a horizontal asymptote at $y = 0$ for $\text{cosec}^{-1} x$.

$\bullet$ The $x$-axis contains values in $(-\infty, -1]$ and $[1, \infty)$.

$\bullet$ The $y$-axis contains angles in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ excluding $0$.

5. Combined View and Reflectional Symmetry

The reflective symmetry about $y = x$ ensures that if $(a, b)$ lies on the cosecant curve, $(b, a)$ lies on the inverse cosecant curve. We can see this in the standard values below:

Point on $\text{cosec } x$ ($a, b$)	Point on $\text{cosec}^{-1} x$ ($b, a$)
$\left( \frac{\pi}{2}, 1 \right)$	$(1, \frac{\pi}{2})$
$\left( \frac{\pi}{6}, 2 \right)$	$(2, \frac{\pi}{6})$
$\left( -\frac{\pi}{6}, -2 \right)$	$(-2, -\frac{\pi}{6})$
$\left( -\frac{\pi}{2}, -1 \right)$	$(-1, -\frac{\pi}{2})$

6. Mathematical Proof of Symmetry

The identity $f(f^{-1}(x)) = x$ confirms the symmetry. For the cosecant function:

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Important Properties of the Inverse Cosecant Function

The properties of $y = \text{cosec}^{-1} x$ mirror those of $\sin^{-1} x$ due to their reciprocal nature, specifically regarding parity and monotonicity.

I. Strictly Decreasing Nature

Unlike the inverse sine function, $y = \text{cosec}^{-1} x$ is strictly decreasing on both segments of its domain. As $x$ increases, the corresponding angle $y$ decreases.

II. Odd Function Property

The inverse cosecant function is an odd function, making it symmetric about the origin.

Derivation:

Let $\text{cosec}^{-1}(-x) = \theta$. This implies $\text{cosec } \theta = -x$, where $\theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\}$.

$x = -\text{cosec } \theta$

Using the identity $\text{cosec}(-\theta) = -\text{cosec } \theta$:

$x = \text{cosec}(-\theta)$

Taking the inverse cosecant of both sides:

$\text{cosec}^{-1} x = -\theta$

Substituting $\theta = \text{cosec}^{-1}(-x)$:

III. Self-Cancellation (Composition Identities)

1. Case of $\text{cosec}(\text{cosec}^{-1} x)$

2. Case of $\text{cosec}^{-1}(\text{cosec } x)$

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Answer:

Properties of Inverse Trigonometric Functions

In this section, we shall explore the essential properties of inverse trigonometric functions. These properties are valid within the Principal Value Branches of the respective functions and are fundamental for solving trigonometric equations and calculus problems.

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Self-Cancellation Properties of Inverse Trigonometric Functions

The Self-Cancellation Properties (also known as Composition Identities) define the behavior of a trigonometric function when it is composed with its own inverse. These identities are fundamental to simplifying expressions and are divided into two distinct forms based on the order of the functions. The validity of these identities depends strictly on the Domain and Range constraints of the functions involved.

Form I: Trigonometric Function of an Inverse ($f(f^{-1}(x)) = x$)

In this form, the inverse trigonometric function is the inner function, and the trigonometric function is the outer function. This composition always results in $x$, provided that $x$ lies within the Domain of the inverse function.

General Rule and Table

For any trigonometric function $f$ and its inverse $f^{-1}$, the identity is given by:

The conditions for all six functions are summarized in the table below:

Function Composition	Condition (Domain of $f^{-1}$)
$\sin (\sin^{-1} x) = x$	$x \in [-1, 1]$ or $|x| \leq 1$
$\cos (\cos^{-1} x) = x$	$x \in [-1, 1]$ or $|x| \leq 1$
$\tan (\tan^{-1} x) = x$	$x \in \mathbb{R}$
$\cot (\cot^{-1} x) = x$	$x \in \mathbb{R}$
$\sec (\sec^{-1} x) = x$	$x \in (-\infty, -1] \cup [1, \infty)$
$\text{cosec } (\text{cosec}^{-1} x) = x$	$x \in (-\infty, -1] \cup [1, \infty)$

Proof of Form I (Example: Sine)

To prove $\sin(\sin^{-1} x) = x$:

Substituting the value of $\theta$ from equation (i) into the equation above:

Form II: Inverse Function of a Trigonometric Function ($f^{-1}(f(x)) = x$)

In this form, the trigonometric function is the inner function and the inverse is the outer function. This identity is not always true for every real number $x$. It is only valid when $x$ lies within the Principal Value Branch (Range) of the inverse function.

General Rule and Table

The identity is expressed as:

The conditions for validity are as follows:

Function Composition	Condition (Range of $f^{-1}$)
$\sin^{-1} (\sin x) = x$	$x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
$\cos^{-1} (\cos x) = x$	$x \in [0, \pi]$
$\tan^{-1} (\tan x) = x$	$x \in (-\frac{\pi}{2}, \frac{\pi}{2})$
$\cot^{-1} (\cot x) = x$	$x \in (0, \pi)$
$\sec^{-1} (\sec x) = x$	$x \in [0, \pi] - \{ \frac{\pi}{2} \}$
$\text{cosec}^{-1} (\text{cosec } x) = x$	$x \in [-\frac{\pi}{2}, \frac{\pi}{2}] - \{ 0 \}$

The Importance of Range Restriction

Trigonometric functions are periodic. If we do not restrict $x$ to the Principal Value Branch, the expression $f^{-1}(f(x))$ will return a value within the principal branch that yields the same trigonometric ratio, rather than the original $x$.

Example: Consider $\sin^{-1}(\sin \frac{2\pi}{3})$. Here $x = \frac{2\pi}{3}$, which is not in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, $\sin^{-1}(\sin \frac{2\pi}{3}) \neq \frac{2\pi}{3}$. The identity only "cancels" when the angle is already within its standard defined boundaries.

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Comparison of Constraints

It is crucial to distinguish between the constraints of Form I and Form II. The comparison is provided in the horizontal table below:

Feature	Form I: $f(f^{-1}(x))$	Form II: $f^{-1}(f(x))$
Result	$x$	$x$
Constraint Type	Domain of Inverse Function	Range of Inverse Function
Nature of $x$	A numerical ratio or value	An angle (in radians)

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Answer:

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Answer:

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Negative Argument Properties of Inverse Trigonometric Functions

The Negative Argument Properties define how inverse trigonometric functions behave when their input value ($x$) is negative. Because the Principal Value Branches (Ranges) are restricted differently for various functions, they are categorized into two distinct groups based on their symmetry. Understanding these properties is vital for evaluating inverse trigonometric expressions accurately without graphical aid.

Group A: Odd Function Symmetry

The functions in Group A have principal value branches that are symmetric about the origin, typically ranging from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Mathematically, these functions satisfy the condition of an odd function, where $f(-x) = -f(x)$.

Properties:

Proof of $\sin^{-1} (-x) = -\sin^{-1} x$:

Group B: Supplementary Symmetry

The functions in Group B have principal value branches that range from $0$ to $\pi$. Since these ranges do not include negative angles, the inverse of a negative ratio cannot simply be the negative of the inverse. Instead, the result is the supplementary angle (mapped into the second quadrant).

Properties:

Detailed Proof of $\cos^{-1} (-x) = \pi - \cos^{-1} x$:

Using the trigonometric identity for the second quadrant, where cosine is negative:

Since $0 \leq y \leq \pi$, the angle $(\pi - y)$ also lies within the interval $[0, \pi]$. Thus, we can apply the inverse function:

Substituting the original value of $y$ from equation (ii):

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Comparative Summary Table

Group Type	Functions	Negative Argument Property	Range Type
Group A (Odd)	$\sin^{-1}, \tan^{-1}, \text{cosec}^{-1}$	$f(-x) = -f(x)$	$[-\frac{\pi}{2}, \frac{\pi}{2}]$
Group B (Supplementary)	$\cos^{-1}, \sec^{-1}, \cot^{-1}$	$f(-x) = \pi - f(x)$	$[0, \pi]$

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Answer:

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Reciprocal Identities of Inverse Trigonometric Functions

The Reciprocal Identities allow us to express inverse trigonometric functions in terms of their reciprocal counterparts. These are particularly useful in calculus and algebraic simplification because most scientific calculators and standard formulas prioritize $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$. By using these identities, we can convert $\text{cosec}^{-1}$, $\sec^{-1}$, and $\cot^{-1}$ into more manageable forms.

Identity for Cosecant and Secant

For the reciprocal functions of sine and cosine, the identity is straightforward because their principal value branches (ranges) are largely overlapping. The only requirement is that $x$ must be within the valid domain of the respective function.

(i) Inverse Cosecant to Inverse Sine

Proof:

(ii) Inverse Secant to Inverse Cosine

Proof:

The Conditional Identity for Inverse Cotangent

The reciprocal identity for $\cot^{-1} x$ is unique because the principal value branch of $\cot^{-1} x$ is $(0, \pi)$, while for $\tan^{-1} x$ it is $(-\frac{\pi}{2}, \frac{\pi}{2})$. This necessitates a case-based formula.

$\cot^{-1} x = \begin{cases} \tan^{-1} \left( \frac{1}{x} \right) & , & x > 0 \\ \pi + \tan^{-1} \left( \frac{1}{x} \right) & , & x < 0 \end{cases}$

Case I: For Positive Arguments ($x > 0$)

When the input $x$ is positive, both $\cot^{-1} x$ and $\tan^{-1} \left( \frac{1}{x} \right)$ yield angles in the First Quadrant. Therefore, the relationship is a direct reciprocal.

Derivation for $x > 0$:

Since $x > 0$ and the range of $\cot^{-1} x$ is $(0, \pi)$, the angle $y$ must lie in the first quadrant, i.e., $y \in (0, \frac{\pi}{2})$.

Now, since $y \in (0, \frac{\pi}{2})$, it falls within the principal value branch of the inverse tangent function $(-\frac{\pi}{2}, \frac{\pi}{2})$. Thus, we can safely write:

Substituting the value of $y$ from equation (i):

Case II: For Negative Arguments ($x < 0$)

When $x < 0$, $\cot^{-1} x$ produces an angle in the Second Quadrant $(\frac{\pi}{2}, \pi)$. However, for a negative argument, $\tan^{-1} \left( \frac{1}{x} \right)$ produces an angle in the Fourth Quadrant $(-\frac{\pi}{2}, 0)$. To equate them, we must adjust the quadrant.

Derivation for $x < 0$:

Since $x < 0$ and $y \in (0, \pi)$, the angle $y$ must lie in the second quadrant, i.e., $y \in (\frac{\pi}{2}, \pi)$.

Let $\tan^{-1} \left( \frac{1}{x} \right) = \theta$. Since $\frac{1}{x}$ is negative, $\theta$ lies in the fourth quadrant, i.e., $\theta \in (-\frac{\pi}{2}, 0)$.

We know that the tangent function has a period of $\pi$, so $\tan y = \tan(y - \pi)$.

If $y \in (\frac{\pi}{2}, \pi)$, then subtracting $\pi$ from the range gives $(y - \pi) \in (-\frac{\pi}{2}, 0)$. This new angle $(y - \pi)$ now falls perfectly within the principal value branch of $\tan^{-1}$.

Substituting the value of $y$ from equation (ii):

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Summary Table

Condition	Identity	Quadrant of $\cot^{-1} x$
$x > 0$	$\cot^{-1} x = \tan^{-1} \left( \frac{1}{x} \right)$	Quadrant I $(0, \frac{\pi}{2})$
$x < 0$	$\cot^{-1} x = \pi + \tan^{-1} \left( \frac{1}{x} \right)$	Quadrant II $(\frac{\pi}{2}, \pi)$

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Answer:

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Complementary Identities of Inverse Trigonometric Functions

In trigonometry, the prefix "co-" in functions like cosine, cotangent, and cosecant stands for complementary. This relationship carries over to inverse trigonometric functions, where the sum of a function and its corresponding "co-function" inverse always results in a constant value of $\frac{\pi}{2}$.

These identities are valid for all values of $x$ that lie within the common domain of the two involved functions. They are extremely powerful for converting one inverse function into another and for simplifying calculus expressions involving derivatives and integrals.

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1. Sine and Cosine Pair

Derivation:

Since the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

For the range of $y$, we check the range of $(\frac{\pi}{2} - y)$:

$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \implies \frac{\pi}{2} \geq -y \geq -\frac{\pi}{2}$

Adding $\frac{\pi}{2}$ to all sides: $\pi \geq \frac{\pi}{2} - y \geq 0$. This falls within the principal branch of $\cos^{-1}$.

Substituting the value of $y$ from equation (i):

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2. Tangent and Cotangent Pair

Derivation:

Checking the range: $-\frac{\pi}{2} < y < \frac{\pi}{2} \implies \frac{\pi}{2} > -y > -\frac{\pi}{2}$.

Adding $\frac{\pi}{2}$ gives $0 < \frac{\pi}{2} - y < \pi$, which is the principal branch $(0, \pi)$ of $\cot^{-1}$.

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3. Secant and Cosecant Pair

Derivation:

Checking the range: $0 \leq y \leq \pi$ and $y \neq \frac{\pi}{2} \implies 0 \geq -y \geq -\pi$ and $-y \neq -\frac{\pi}{2}$.

Adding $\frac{\pi}{2}$ gives $\frac{\pi}{2} \geq \frac{\pi}{2} - y \geq -\frac{\pi}{2}$ and $(\frac{\pi}{2} - y) \neq 0$. This matches the branch of $\text{cosec}^{-1}$.

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Summary of Domain Constraints

It is essential to remember that these identities only hold when $x$ is in the valid domain for both functions in the pair.

Identity Pair	Valid Domain ($x$)
$(\sin^{-1} x, \cos^{-1} x)$	$[-1, 1]$
$(\tan^{-1} x, \cot^{-1} x)$	$\mathbb{R}$
$(\sec^{-1} x, \text{cosec}^{-1} x)$	$(-\infty, -1] \cup [1, \infty)$

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Answer:

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Sum and Difference of Inverse Tangent Functions

The addition and subtraction formulas for inverse tangent functions are among the most frequently used identities in trigonometry, calculus, and coordinate geometry. These identities allow us to combine multiple inverse tangent terms into a single term. However, unlike basic algebraic addition, these formulas depend strictly on the values of the arguments $x$ and $y$ to ensure the result stays within the Principal Value Branch $(-\frac{\pi}{2}, \frac{\pi}{2})$.

1. Addition Formula for $\tan^{-1} x + \tan^{-1} y$

The general form of the addition formula is derived from the compound angle formula: $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$. The resulting identity changes based on the product $xy$.

Case I: When $xy < 1$

This is the standard case where the sum of the angles remains within the principal range.

Derivation:

Using the compound angle formula:

When $xy < 1$, the value of $(\alpha + \beta)$ falls within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. Therefore, we can directly apply the inverse function:

Case II: When $xy > 1$ and $x, y > 0$

If $x$ and $y$ are both positive and their product exceeds $1$, the sum of the angles $\alpha + \beta$ exceeds $\frac{\pi}{2}$. To bring the result back into the principal range, we must add $\pi$.

Derivation:

Let $\tan^{-1} x = \alpha$ and $\tan^{-1} y = \beta$. Since $x, y > 0$, $\alpha$ and $\beta$ are in $(0, \frac{\pi}{2})$.

If $xy > 1$, then $\tan \alpha \tan \beta > 1$, which implies $\alpha + \beta > \frac{\pi}{2}$. However, since both are less than $\frac{\pi}{2}$, their sum is less than $\pi$. Thus, $(\alpha + \beta) \in (\frac{\pi}{2}, \pi)$.

The angle $(\alpha + \beta)$ is outside the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$. We use the property $\tan(\theta - \pi) = \tan \theta$:

Since $\alpha + \beta \in (\frac{\pi}{2}, \pi)$, it follows that $(\alpha + \beta - \pi) \in (-\frac{\pi}{2}, 0)$. This value is now in the principal range.

Case III: When $xy > 1$ and $x, y < 0$

For negative $x$ and $y$ with $xy > 1$, the sum $\alpha + \beta$ will be less than $-\frac{\pi}{2}$. In this case, we subtract $\pi$.

2. Subtraction Formula for $\tan^{-1} x - \tan^{-1} y$

The subtraction formula is more straightforward because the denominator $(1 + xy)$ usually keeps the result within the principal range for most standard domains.

Derivation:

Using the identity $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$:

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Summary Table of Formulas

Operation	Condition	Resulting Formula
Addition	$xy < 1$	$\tan^{-1} \left( \frac{x+y}{1-xy} \right)$
Addition	$xy > 1, x>0, y>0$	$\pi + \tan^{-1} \left( \frac{x+y}{1-xy} \right)$
Addition	$xy > 1, x < 0, y < 0$	$-\pi + \tan^{-1} \left( \frac{x+y}{1-xy} \right)$
Subtraction	$xy > -1$	$\tan^{-1} \left( \frac{x-y}{1+xy} \right)$

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Answer:

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Conversion of $2\tan^{-1} x$

The term $2\tan^{-1} x$ is one of the most versatile expressions in inverse trigonometry. It can be transformed into Inverse Sine, Inverse Cosine, or Inverse Tangent depending on the requirement of the mathematical problem. These conversions are fundamentally derived from the double angle trigonometric identities of $\sin 2\theta$, $\cos 2\theta$, and $\tan 2\theta$ expressed in terms of $\tan \theta$.

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1. Conversion into Sine Inverse Form

This conversion is used when we need to simplify an expression involving $\sin^{-1}$. It is valid only when the resulting argument lies within the domain $[-1, 1]$.

The Identity:

Derivation:

We know the standard trigonometric identity for $\sin 2y$ in terms of $\tan y$:

Substituting $x$ for $\tan y$ in the above equation from (i):

Taking inverse sine on both sides:

Replacing $y$ with $\tan^{-1} x$:

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2. Conversion into Cosine Inverse Form

This form is particularly useful when dealing with square terms and wanting to eliminate radicals in calculus or algebraic simplifications.

The Identity:

Derivation:

We utilize the double angle identity for cosine:

Substituting the value of $x$ from (ii):

Substituting back $y = \tan^{-1} x$:

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3. Conversion into Tangent Inverse Form

This conversion is strictly used to combine or reduce multiple $\tan^{-1}$ terms into a single expression.

The Identity:

Derivation:

We use the double angle identity: $\tan 2y = \frac{2\tan y}{1 - \tan^2 y}$.

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Summary of Conversion Formulas

Target Function	Equivalent Formula	Domain Constraint
$\sin^{-1}$	$\sin^{-1} \left( \frac{2x}{1+x^2} \right)$	$-1 \leq x \leq 1$
$\cos^{-1}$	$\cos^{-1} \left( \frac{1-x^2}{1+x^2} \right)$	$x \geq 0$
$\tan^{-1}$	$\tan^{-1} \left( \frac{2x}{1-x^2} \right)$	$-1 < x < 1$

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Answer:

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Triple Angle Inverse Identities

The Triple Angle Inverse Identities allow us to express three times an inverse trigonometric function as a single inverse function. These identities are the inverse counterparts of the standard trigonometric triple angle formulas ($\sin 3\theta, \cos 3\theta, \text{ and } \tan 3\theta$). They are primarily utilized in simplifying complex inverse trigonometric expressions and are highly relevant for the CBSE Class 12 curriculum and competitive examinations like JEE Mains.

Identity for $3\sin^{-1} x$

This identity is used to consolidate the term $3\sin^{-1} x$ into a single inverse sine function. It is valid only when the resulting angle remains within the principal value branch $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

The Formula:

Derivation:

We know from the standard triple angle formula for sine:

Substituting $x = \sin \theta$ into the right-hand side:

Taking the inverse sine of both sides:

Replacing $\theta$ with $\sin^{-1} x$:

Identity for $3\cos^{-1} x$

Similar to the sine identity, this is derived from the $\cos 3\theta$ formula. Its validity is restricted to values of $x$ that keep the angle within $[0, \pi]$.

The Formula:

Derivation:

Using the standard triple angle formula for cosine:

Substituting $x = \cos \theta$:

Identity for $3\tan^{-1} x$

This identity is used to combine three inverse tangent terms. It is defined within an interval where the denominator $1 - 3x^2$ does not result in an undefined value.

The Formula:

Derivation:

Using the standard triple angle formula for tangent:

Substituting $x = \tan \theta$:

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Summary of Triple Angle Constraints

The constraints on $x$ are as important as the formula itself. They ensure that the resulting angle falls within the Principal Value Branch of the inverse function.

Left Hand Side	Right Hand Side Equivalent	Valid Domain of $x$
$3\sin^{-1} x$	$\sin^{-1} (3x - 4x^3)$	$x \in [-\frac{1}{2}, \frac{1}{2}]$
$3\cos^{-1} x$	$\cos^{-1} (4x^3 - 3x)$	$x \in [\frac{1}{2}, 1]$
$3\tan^{-1} x$	$\tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$	$x \in (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$

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Answer:

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Conversion of $\sin^{-1} x$ and $\cos^{-1} x$ into $\tan^{-1}$ Form

In Inverse Trigonometry, it is often necessary to convert $\sin^{-1} x$ or $\cos^{-1} x$ into the $\tan^{-1}$ form. This is particularly useful because the addition and subtraction formulas for $\tan^{-1}$ are algebraically simpler to handle than those for sine or cosine inverse. These conversions are based on the fundamental trigonometric identities relating Sine, Cosine, and Tangent.

Conversion of $\sin^{-1} x$ to $\tan^{-1}$

This conversion expresses an inverse sine function in terms of an inverse tangent function by considering the ratio of the "perpendicular" to the "base" in a right-angled triangle.

The Identity:

Derivation:

We know the identity relating $\cos \theta$ and $\sin \theta$:

Now, we express $\tan \theta$ as the ratio of sine and cosine:

Substituting the value of $\theta$ from equation (i):

Conversion of $\cos^{-1} x$ to $\tan^{-1}$

Similar to the sine conversion, the inverse cosine can be converted by identifying the "base" and "hypotenuse" relationship.

The Identity:

Derivation:

We know the identity for $\sin \phi$:

Now, expressing $\tan \phi$:

Substituting the value of $\phi$ from equation (ii):

Quick Conversion Table (Triangle Method)

For any inverse trigonometric function, assume a right-angled triangle where the argument represents the ratio of two sides. Let P be Perpendicular, B be Base, and H be Hypotenuse.

Function	Assume Ratio	Missing Side (by Pythagoras)	$\tan^{-1}$ Equivalent
$\sin^{-1} x$	$P=x, H=1$	$B = \sqrt{1-x^2}$	$\tan^{-1}(\frac{P}{B}) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
$\cos^{-1} x$	$B=x, H=1$	$P = \sqrt{1-x^2}$	$\tan^{-1}(\frac{P}{B}) = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$
$\text{cosec}^{-1} x$	$H=x, P=1$	$B = \sqrt{x^2-1}$	$\tan^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)$

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Answer:

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Inter-conversion of Sine and Cosine Inverse Functions

In inverse trigonometry, Sine Inverse and Cosine Inverse are closely related through the Pythagorean identity. Since $\sin^2 \theta + \cos^2 \theta = 1$, we can express one inverse function in terms of the other. This is particularly useful when solving equations where different inverse functions are present or when we need to find the value of a trigonometric function composed with a different inverse function.

Conversion of $\sin^{-1} x$ into $\cos^{-1}$

This conversion allows us to switch from a sine-based inverse to a cosine-based inverse using the square root relationship derived from the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$.

The Identity:

Derivation:

We know that $\cos^2 \theta + \sin^2 \theta = 1$. Therefore:

Substituting $\theta$ from equation (i):

Conversion of $\cos^{-1} x$ into $\sin^{-1}$

Similarly, we can express cosine inverse in terms of sine inverse using $\sin \theta = \sqrt{1 - \cos^2 \theta}$.

The Identity:

Derivation:

Composition Property: $\cos(\sin^{-1} x)$ and $\sin(\cos^{-1} x)$

A very important result in trigonometry is that the value of cosine of sine inverse and sine of cosine inverse is exactly the same for any $x$ in the domain $[-1, 1]$.

Fundamental Result:

Proof for $\cos(\sin^{-1} x)$:

Proof for $\sin(\cos^{-1} x)$:

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Answer:

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Summary Table of Relationships

Function Composition	Resulting Expression	Domain
$\sin(\cos^{-1} x)$	$\sqrt{1 - x^2}$	$[-1, 1]$
$\cos(\sin^{-1} x)$	$\sqrt{1 - x^2}$	$[-1, 1]$
$\sin^{-1} x$	$\cos^{-1} \sqrt{1 - x^2}$	$[0, 1]$
$\cos^{-1} x$	$\sin^{-1} \sqrt{1 - x^2}$	$[0, 1]$

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Addition and Subtraction Formulas for Inverse Sine and Cosine

The addition and subtraction formulas for Inverse Sine and Inverse Cosine allow us to combine two different inverse trigonometric terms into a single term. These identities are derived directly from the compound angle formulas of Sine and Cosine ($\sin(A \pm B)$ and $\cos(A \pm B)$). These are extensively used in solving trigonometric equations and simplifying complex calculus problems.

Formulas for Inverse Sine

For $x, y \in [-1, 1]$, the formulas for addition and subtraction of $\sin^{-1}$ are as follows:

(i) Addition: $\sin^{-1} x + \sin^{-1} y$

(ii) Subtraction: $\sin^{-1} x - \sin^{-1} y$

To derive these formulas, we use the substitution method based on standard trigonometric compound angle identities.

Using the Pythagorean identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$:

For Addition ($\sin^{-1} x + \sin^{-1} y$):

We know that $\sin(A + B) = \sin A \cos B + \cos A \sin B$. Substituting the values:

For Subtraction ($\sin^{-1} x - \sin^{-1} y$):

We know that $\sin(A - B) = \sin A \cos B - \cos A \sin B$. Substituting the values:

Formulas for Inverse Cosine

For $x, y \in [-1, 1]$, the formulas for addition and subtraction of $\cos^{-1}$ are as follows:

(i) Addition: $\cos^{-1} x + \cos^{-1} y$

(ii) Subtraction: $\cos^{-1} x - \cos^{-1} y$

Using the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$:

For Addition ($\cos^{-1} x + \cos^{-1} y$):

Using $\cos(A + B) = \cos A \cos B - \sin A \sin B$:

For Subtraction ($\cos^{-1} x - \cos^{-1} y$):

Using $\cos(A - B) = \cos A \cos B + \sin A \sin B$:

Summary of Sign Conventions

Note the sign convention in these formulas: inverse sine addition results in addition in the argument, whereas inverse cosine addition results in subtraction in the argument.

Function Operation	Argument Operator	Equivalent Formula
$\sin^{-1} x + \sin^{-1} y$	$+$	$\sin^{-1} \left( x\sqrt{1-y^2} + y\sqrt{1-x^2} \right)$
$\sin^{-1} x - \sin^{-1} y$	$-$	$\sin^{-1} \left( x\sqrt{1-y^2} - y\sqrt{1-x^2} \right)$
$\cos^{-1} x + \cos^{-1} y$	$-$	$\cos^{-1} \left( xy - \sqrt{1-x^2}\sqrt{1-y^2} \right)$
$\cos^{-1} x - \cos^{-1} y$	$+$	$\cos^{-1} \left( xy + \sqrt{1-x^2}\sqrt{1-y^2} \right)$

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Answer: