θ principal branch
sin⁻¹
Chapter 2  ·  Class XII Mathematics

Choosing the One True Branch

Inverse Trigonometric Functions

Tame the Multi-Valued Beast — Master Principal Values and Their Identities

📖 Read Full Notes ⚡ Formula Capsules

Chapter Snapshot

8Concepts
14Formulae
4–6%Exam Weight
2–3Avg Q's
ModerateDifficulty

Why This Chapter Matters for Entrance Exams

CBSEJEE MainJEE Advanced

JEE Main frequently tests domain/range and principal-value simplification, while JEE Advanced favours multi-step identity chains combining two or three inverse functions. CBSE Boards set at least one 2-mark simplification and one identity-proof question every year.

Key Concept Highlights

Principal Value Branch
Domain and Range of Each Inverse Trig Function
Graphs of Inverse Trig Functions
Complementary Function Identities
Sum and Difference Formulae
Conversion Between Inverse Functions
Double-Angle Inverse Formulae
Simplification Using Substitution (x = tanθ, sinθ, etc.)

Important Formula Capsules

$\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$
$\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$
$\sec^{-1}x + \text{cosec}^{-1}x = \dfrac{\pi}{2}$
$\sin^{-1}(-x) = -\sin^{-1}x$
$\tan^{-1}(-x) = -\tan^{-1}x,\quad \cos^{-1}(-x) = \pi - \cos^{-1}x$
$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right),\ xy<1$
$2\tan^{-1}x = \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right),\ |x|\le 1$
$2\tan^{-1}x = \cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right),\ x\ge 0$

What You Will Learn

Navigate to Chapter Resources

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🏆 Exam Strategy & Preparation Tips

Build a single reference table of domain/range for all six functions and revise it daily — most errors come from forgetting the restricted range, not the algebra. For JEE Advanced, practise substitution tricks (x = tanθ, x = sinθ) on 15–20 problems. Time investment: 2 days.

Chapter 1 · CBSE · Class XII
📐

Inverse of Trigonometric Functions

Inverse Trigonometric Functions NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Principal Values Inverse Functions Trigonometric Identities Domain and Range JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
15 sections
🗺️ Overview
Inverse trigonometric functions are obtained by reversing suitably restricted trigonometric functions. Since ordinary trigonometric functions are periodic, they are not one-one over their entire domain. Therefore, before defining an inverse, we restrict the domain of the trigonometric function so that every element of the range corresponds to exactly one element of the domain.

This chapter is one of the most important chapters of Class XII Mathematics because it provides the foundation for differentiation of inverse trigonometric functions, integration, limits, continuity, differential equations, vectors, three-dimensional geometry and numerous applications in Physics and Engineering.
🤔 Did You Know?
Why Do We Need Inverse Trigonometric Functions?
Suppose we know that
\[\sin \theta=\frac{3}{5}\]
The natural question is:
What is the value of \(\theta\)?
To answer such questions mathematically, we require the inverse of the sine function. Similarly,
  • If \(\cos \theta=a\), then we use \(\cos^{-1}a\).
  • If \(\tan \theta=a\), then we use \(\tan^{-1}a\).
  • If \(\cot \theta=a\), then we use \(\cot^{-1}a\).
  • If \(\sec \theta=a\), then we use \(\sec^{-1}a\).
  • If \(\operatorname{cosec} \theta=a\), then we use \(\operatorname{cosec}^{-1}a\).
Thus, inverse trigonometric functions determine the angle corresponding to a known trigonometric ratio.
🗒️ Why is Domain Restriction Necessary?
Why is Domain Restriction Necessary?
Every inverse function exists only if the original function is one-one and onto. Unfortunately, none of the six trigonometric functions is one-one over the entire set of real numbers because of periodicity.
For example,
\[\sin 30^\circ=\sin 150^\circ=\sin 390^\circ=\frac12\]
One output corresponds to infinitely many inputs. Hence, the sine function is not one-one. Therefore, its inverse cannot be defined unless we restrict its domain. Similar situations occur for all trigonometric functions.
📌 Range of Trigonometric Functions
📎 Important Note
  • The above are ranges, not domains.
  • Students often confuse excluded domain values with the range.
  • \(\tan x\) and \(\cot x\) can produce every real number.
  • \(\sec x\) and \(\operatorname{cosec} x\) never take values between -1 and 1.
🗒️ Inverse Function
Suppose \(f:X\rightarrow Y\) is one-one and onto. Then there exists a unique function \(g:Y\rightarrow X\) such that
\[g(y)=x\]
where
\(y=f(x)\)

The function \(g\) is called the inverse of \(f\) and is denoted by
\[g=f^{-1}\]
Therefore,
  • Domain of \(f^{-1}\) = Range of \(f\)
  • Range of \(f^{-1}\) = Domain of \(f\)
🏷️ Fundamental Properties of an Inverse Function
If \(g=f^{-1}\) then
\[(f^{-1}\circ f)(x)=x\]
and
\[(f\circ f^{-1})(y)=y\]
Thus,
\[f^{-1}(f(x))=x\]
and
\[f(f^{-1}(y))=y\]
Also,
\[(f^{-1})^{-1}=f\]
Hence the inverse of the inverse is the original function.
📐 Derivation of Composition Property
Suppose \(y=f(x)\). Applying the inverse function,
\[f^{-1}(y)=x\]
Since \(y=f(x)\).
therefore,
\[f^{-1}(f(x))=x\]
Similarly,
\[x=f^{-1}(y)\]
Applying the original function, \(f(x)=y\)
Hence,
\[f(f^{-1}(y))=y\]
These two identities are the basis of every inverse function.
💡 Concept of Principal Branch
🗺️ Roadmap for Solving Inverse Trigonometric Problems

  1. Identify the trigonometric ratio.

  2. Check whether its value belongs to the permissible range.

  3. Use the appropriate inverse function.

  4. Ensure that the obtained angle belongs to the principal value interval.

  5. Simplify using standard angles whenever possible.

✏️ Example
Solved Example
1
Question
Find \(\sin^{-1}\left(\frac12\right)\)
  • Principal value of inverse sine
  • Standard trigonometric values
  1. Since
    \[\sin\frac{\pi}{6}=\frac12\]
  2. and
    \[ \frac{\pi}{6}\in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \]
  3. therefore,
    \[ \boxed{ \sin^{-1}\left(\frac12\right)=\frac{\pi}{6} } \]
2
Question
Find \(\cos^{-1}(1)\)
  1. Since
    \[\cos0=1\]
  2. and
    \[0\in[0,\pi]\]
  3. therefore,
    \[\boxed{\cos^{-1}(1)=0}\]
3
Question
FInd \(\tan^{-1}(-1)\)
🗒️ Solurion

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🔢 Quick Formula Summary
🌟 Board and Competitive Exam Significance
  • Foundation for differentiation of inverse trigonometric functions.
  • Frequently used in definite and indefinite integration.
  • Appears in JEE Main, JEE Advanced, NDA, CUET and state engineering entrance examinations.
  • Essential for solving trigonometric equations.
  • Applied in coordinate geometry, vectors and 3D geometry.
  • Widely used in Physics while resolving angles from measured ratios.
❌ Common Mistakes
  • Writing \(\sin^{-1}x\) as \(\dfrac1{\sin x}\). It actually denotes the inverse function.
  • Ignoring principal value intervals.
  • Confusing the range of a trigonometric function with its domain.
  • Using unrestricted angles while evaluating inverse trigonometric functions.
  • Assuming every function automatically possesses an inverse.
🗒️ CBSE Case Study (HOTS)

Situation

A surveyor measures the height of a tower by observing that

\[\sin\theta=\frac45\]

Instead of solving a triangle repeatedly, he directly calculates

\[\theta=\sin^{-1}\left(\frac45\right).\]

Questions

  1. Why is the inverse sine function used?
  2. Why is the obtained angle unique?
  3. What restriction on the sine function guarantees uniqueness?

Learning Outcome

Students understand the practical necessity of inverse trigonometric functions and the roleplayed by principal branches in producing a unique angle.

📐

Inverse Sine Function (Principal Branch of Sine Function)

📋
Table of Contents
13 sections
🗺️ Overview
The sine function is one of the six basic trigonometric functions and is defined for every real number. Its domain is the set of all real numbers, whereas its range is the closed interval
\[ [-1,1]. \]
Since every value in the interval \([-1,1]\) is attained infinitely many times due to the periodic nature of the sine function, the function is not one-one over its entire domain. Consequently, an inverse function cannot be defined unless the domain is suitably restricted.
📌 Note
🗒️ Restriction Of Domain
To define the inverse of the sine function, we restrict its domain to an interval over which it becomes both one-one and onto.
The standard (principal) restriction is
\[\boxed{\bbox[2pt]{-\frac{\pi}{2}\le x\le\frac{\pi}{2}}}\]
On this interval,
  • the sine function is strictly increasing,
  • every value from \(-1\) to \(1\) occurs exactly once,
  • therefore it becomes one-one and onto.
Hence,
\[ \sin: \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow [-1,1] \]
is invertible.
🎨 SVG Diagram
Principal Branch
+1 −1 −π −π/2 0 π/2 π (−π/2, −1) (π/2, 1) x y = sin(x) PRINCIPAL BRANCH: [−π/2, π/2]
📎 Other Possible Restricted Domains
The sine function also becomes one-one on several other intervals of length \(\pi\). Some examples are:
\[ \left[ -\frac{3\pi}{2}, -\frac{\pi}{2} \right], \quad \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], \quad \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]. \]
Each of these intervals has the same range
\[[-1,1]\]
Therefore, an inverse function can theoretically be defined on each of these intervals. Each such inverse is called a branch of the inverse sine function.
📌 Principal Branch of the Inverse Sine Function
🎨 SVG Diagram
Graphical representation of \(\sin^{-1} x\)
Y Y′ X X′ O −1 1 5π 2 2π 3π 2 π π 2 −π 2 −π −3π 2 −2π −5π 2 y = sin⁻¹ x
📘 Definition of the Inverse Sine Function
🎨 Principal Branch of \(\sin^{-1}\)
π π/2 0 −π/2 −π −1 +1 (−1, −π/2) (1, π/2) x y = sin⁻¹(x) PRINCIPAL BRANCH: y = Sin⁻¹(x)
🔢 Fundamental Identities
🤔 Why is the Second Identity Restricted?
Consider
\[\sin^{-1}\left(\sin\frac{5\pi}{6}\right).\]
Since
\[\sin\frac{5\pi}{6}=\frac12,\]
therefore,
\[\sin^{-1}\left(\frac12\right)=\frac{\pi}{6},\]
not
\[\frac{5\pi}{6}\]
This happens because the inverse sine function always returns the principal value lying in
\[\left[-\frac{\pi}{2},\frac{\pi}{2}\right].\]
🎨 SVG Diagram
Graphical Representation
+1.0 +0.5 0 -0.5 -1.0 -2π -3π/2 -π -π/2 π/2 π 3π/2 2π x (rad) y = sin(x) (π/2, 1) (-3π/2, 1) (3π/2, -1) (-π/2, -1) 1 Full Cycle / Period (2π) SINE GRAPH ACROSS TWO PERIODS (-2π to +2π)
🔍 Graphical Interpretation
The graph of the sine function over the interval
\[\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\]
is strictly increasing. Every horizontal line intersects the graph at exactly one point, satisfying the horizontal line test. Therefore, the function possesses an inverse on this interval.
✏️ Example
Solved Example
1
Question
Evaluate
\[\sin^{-1}(1)\]
  1. 1
    Recall the standard angle whose sine is 1.
  2. 2
    Verify that the angle lies in the principal interval.
\[\sin\frac{\pi}{2}=1\]
Since
\[\frac{\pi}{2}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\]
therefore,
\[\boxed{\sin^{-1}(1)=\frac{\pi}{2}.}\]
2
Question
Evaluate
\[\sin^{-1}\left(-\frac{\sqrt3}{2}\right).\]
Since
\[\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt3}{2},\]
and
\[-\frac{\pi}{3}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\]
therefore,
\[\boxed{\sin^{-1}\left(-\frac{\sqrt3}{2}\right)=-\frac{\pi}{3}}\]
3
Question
Evaluate
\[\sin^{-1}\left(\sin\frac{7\pi}{6}\right)\]
\[\sin\frac{7\pi}{6}=-\frac12\]
Now,
\[\sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}\]
Hence,
\[\boxed{\sin^{-1}\left(\sin\frac{7\pi}{6}\right)=-\frac{\pi}{6}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Confusing inverse with reciprocal.
  • Ignoring the principal value interval.
  • Writing \(\sin^{-1}(\sin x)=x\) for every real number.
  • Using angles outside the permitted range without converting them to principal values.
  • Incorrectly assuming that the inverse sine function has domain \(\mathbb{R}\).
📋 CBSE Case Study (HOTS)

An engineer measures the inclination of a ramp. The ratio of its height to its length is

\[\frac35.\]

Instead of drawing a triangle every time, the calculator computes

\[\theta=\sin^{-1}\left(\frac35\right)\]

Questions

  1. Why is the inverse sine function used instead of the sine function?
  2. Why does the calculator return only one angle?
  3. What is the principal value interval responsible for this uniqueness?
  4. Would the answer change if the principal branch were chosen differently?
📐

Inverse Cosine Function (Principal Branch of Cosine Function)

📋
Table of Contents
13 sections
🗺️ Overview
The cosine function is defined for every real number and has range \([-1,1]\). However, because the cosine function is periodic, each value in its range is attained infinitely many times. Hence, the cosine function is not one-one on the set of all real numbers and therefore does not possess an inverse over its entire domain.

To define the inverse cosine function uniquely, we restrict the domain of the cosine function to an interval where it becomes one-one and onto.
📌 Domain and Range of the Cosine Function
🎨 SVG Diagram
Graphical Representation of Cosine
+1.0 +0.5 0 -0.5 -1.0 -2π -3π/2 -π -π/2 π/2 π 3π/2 2π x (rad) y = cos(x) (-2π, 1) (0, 1) (2π, 1) (-π, -1) (π, -1) 1 Full Period (2π) COSINE GRAPH ACROSS TWO PERIODS (-2π to +2π)
🗒️ Restriction Of Domain
To define an inverse function, the cosine function must be restricted to an interval over which it is one-one and onto. The standard restriction is
\[\boxed{0\le x\le\pi}\]
On this interval,
  • the cosine function is strictly decreasing,
  • every value between 1 and -1 occurs exactly once,
  • hence the function becomes one-one and onto.
Therefore,
\[\cos:[0,\pi]\rightarrow[-1,1]\]
is invertible.
📎 Side Note
Other Possible Restricted Domains
The cosine function is also one-one on many other intervals of length \(\pi\). Some examples are
\[[-\pi,0],\quad[0,\pi],\quad[\pi,2\pi],\quad[-2\pi,-\pi]\]
Each of these intervals has the same range
\[[-1,1]\]
Hence, an inverse function can be defined on each such interval. Every such inverse is called a branch of the inverse cosine function.
📌 Principal Branch of the Inverse Cosine Function
📘 Definition of the Inverse Cosine Function
🎨 SVG Diagram
Graphical Representation of \(\cos^{-1} x\)
Y Y′ X X′ O −1 1 5π 2 2π 3π 2 π π 2 −π 2 −π −3π 2 −2π −5π 2 y = cos⁻¹ x
🔢 Fundamental Identities
🤔 Why is the Second Identity Restricted?
Consider
\[\cos^{-1}\left(\cos\frac{5\pi}{3}\right)\]
Since
\[\cos\frac{5\pi}{3}=\frac12,\]
therefore,
\[\cos^{-1}\left(\frac12\right)=\frac{\pi}{3},\]
not
\[\frac{5\pi}{3}\]
This happens because the inverse cosine function always returns an angle belonging to the principal interval
\[[0,\pi]\]
🎨 SVG Diagram
Principal Branch of Cosine
+1 −1 −π −π/2 0 π/2 π 3π/2 2π (0, 1) (π, −1) x y = cos(x) PRINCIPAL BRANCH: [0, π]
📎 Graphical Interpretation
On the interval \([0,\pi],\) the cosine curve is strictly decreasing.
Every horizontal line intersects the graph at exactly one point, satisfying the horizontal line test. Hence, the cosine function possesses an inverse on this interval.
🎨 SVG Diagram
PRINCIPAL BRANCH: y = Cos⁻¹(x)
3π/2 π π/2 0 −π/2 −1 +1 (−1, π) (0, π/2) (1, 0) x y = cos⁻¹(x) PRINCIPAL BRANCH: y = Cos⁻¹(x)
✏️ Example
Solved Example
1
Question
Evaluate \(\cos^{-1}(1)\)
  1. 1
    Recall the standard angle whose cosine is 1.
  2. 2
    Verify that the angle belongs to the principal interval.
\[\cos0=1\]
Since
\[0\in[0,\pi]\]
therefore
\[\boxed{\cos^{-1}(1)=0}\]
2
Question
Evaluate \(\cos^{-1}\left(-\frac12\right)\)
Since
\[\cos\frac{2\pi}{3}=-\frac12,\]
and
\[\frac{2\pi}{3}\in[0,\pi],\]
therefore
\[\boxed{\cos^{-1}\left(-\frac12\right)=\frac{2\pi}{3}.}\]
3
Question
Evaluate \(\cos^{-1}\left(\cos\frac{5\pi}{4}\right)\)
\[\cos\frac{5\pi}{4}=-\frac{\sqrt2}{2}\]
Now,
\[\cos^{-1}\left(-\frac{\sqrt2}{2}\right)=\frac{3\pi}{4}\]
Hence,
\[\boxed{\cos^{-1}\left(\cos\frac{5\pi}{4}\right)=\frac{3\pi}{4}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\cos^{-1}x=\dfrac1{\cos x}\).
  • Confusing inverse cosine with reciprocal cosine.
  • Ignoring the principal value interval.
  • Using \(\cos^{-1}(\cos x)=x\) for every real number.
  • Choosing an angle outside the interval \([0,\pi]\).
📋 CBSE Case Study (HOTS)

A drone flies towards a building. The horizontal distance from the drone to the building and the direct line-of-sight distance give the ratio

\[\cos\theta=\frac45.\]

The onboard computer immediately calculates

\[\theta=\cos^{-1}\left(\frac45\right).\]

Questions

  1. Why is the inverse cosine function used?
  2. Why does the computer return only one angle?
  3. What is the principal value interval of the inverse cosine function?
  4. Why is the interval \([0,\pi]\) preferred over other possible intervals?
📐

Inverse Cosecant Function (Principal Branch of Cosecant Function)

📋
Table of Contents
13 sections
📘 Definition
📌 Domain and Range of the Cosecant Function
📎 Restriction of Domain
To define the inverse cosecant function uniquely, the domain of the cosecant function is restricted to
\[\boxed{\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}}\]
The point \(0\) is excluded because
\[\sin0=0\]
and therefore the cosecant function is undefined at this point.
On the interval
\[\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}\]
the cosecant function becomes one-one and its range is \((-\infty,-1]\cup[1,\infty)\)
Hence,
\[\operatorname{cosec}:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}\rightarrow(-\infty,-1]\cup[1,\infty)\]
is invertible.
📌 Other Possible Restricted Domains
📎 Principal Branch of the Inverse Cosecant Function
Among all possible branches, the interval
\[\boxed{\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}}\]
is accepted as the principal branch. This convention is followed in NCERT, CBSE, JEE Main, JEE Advanced and most university-level mathematics. Thus, the inverse cosecant function is defined as
\[\boxed{\operatorname{cosec}^{-1}:(-\infty,-1]\cup[1,\infty)\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}}\]
📘 Definition
🔢 Fundamental Identities
🤔 Did You Know?
Why is the Second Identity Restricted?
Consider
\[\operatorname{cosec}^{-1}\left(\operatorname{cosec}\frac{5\pi}{6}\right)\]
Since
\[\operatorname{cosec}\frac{5\pi}{6}=2\]
therefore
\[\operatorname{cosec}^{-1}(2)=\frac{\pi}{6}\]
not
\[\frac{5\pi}{6}\]
This happens because the inverse cosecant function always returns the principal value belonging to
\[\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}\]
🗒️ Graphical Interpretation
On the interval
\[\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\},\]
the cosecant function has two continuous branches separated by the vertical asymptote at \(x=0\) Every value belonging to
\[(-\infty,-1]\cup[1,\infty)\]
is attained exactly once. Therefore, the function satisfies the horizontal line test and possesses a unique inverse.
✏️ Example
Solved Example
1
Question
Evaluate \(\operatorname{cosec}^{-1}(2)\)
  1. 1
    Recall the angle whose cosecant equals 2.
  2. 2
    Check whether it lies in the principal interval.
\[\operatorname{cosec}\frac{\pi}{6}=2\]
Since
\[\frac{\pi}{6}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}\]
therefore,
\[\boxed{\operatorname{cosec}^{-1}(2)=\frac{\pi}{6}}\]
2
Question
Evaluate \(\operatorname{cosec}^{-1}(-2)\)
\[\operatorname{cosec}\left(-\frac{\pi}{6}\right)=-2\]
Therefore,
\[\boxed{\operatorname{cosec}^{-1}(-2)=-\frac{\pi}{6}}\]
3
Question
Evaluate \(\operatorname{cosec}^{-1}\left(\operatorname{cosec}\frac{7\pi}{6}\right)\)
\[\operatorname{cosec}\frac{7\pi}{6}=-2\]
Now,
\[\operatorname{cosec}^{-1}(-2)=-\frac{\pi}{6}\]
Hence,
\[\boxed{\operatorname{cosec}^{-1}\left(\operatorname{cosec}\frac{7\pi}{6}\right)=-\frac{\pi}{6}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\operatorname{cosec}^{-1}x=\dfrac1{\operatorname{cosec} x}\).
  • Including \(0\) in the principal interval.
  • Forgetting that inverse cosecant exists only for \(|x|\ge1\).
  • Using \(\operatorname{cosec}^{-1}(\operatorname{cosec} x)=x\) for every real number.
  • Confusing reciprocal functions with inverse functions.
🗒️ CBSE Case Study (HOTS)

A surveyor measures the ratio of the hypotenuse to the opposite side of a right triangle and obtains

\[\operatorname{cosec}\theta=2.\]

The surveying software immediately computes

\[\theta=\operatorname{cosec}^{-1}(2)\]

Questions

  1. Why is the inverse cosecant function used?
  2. Why is the principal value unique?
  3. Why is \(0\) excluded from the principal interval?
  4. What is the domain of the inverse cosecant function?
📐

Inverse Secant Function (Principal Branch of Secant Function)

📋
Table of Contents
13 sections
📘 Definition
📌 Domain and Range of the Cosecant Function
📎 Restriction of Domain
To define the inverse secant function uniquely, the domain of the secant function is restricted to
\[\boxed{[0,\pi]-\left\{\frac{\pi}{2}\right\}}\]
The value \(\dfrac{\pi}{2}\) is excluded because
\[\cos\frac{\pi}{2}=0,\]
making the secant function undefined.

On the interval
\[[0,\pi]-\left\{\frac{\pi}{2}\right\},\]
the secant function becomes one-one and its range is
\[(-\infty,-1]\cup[1,\infty)\]
Hence,
\[\sec:[0,\pi]-\left\{\frac{\pi}{2}\right\}\rightarrow(-\infty,-1]\cup[1,\infty)\]
is invertible.
📌 Other Possible Restricted Domains
📎 Principal Branch of the Inverse Secant Function
Among all possible branches, the interval
\[\boxed{[0,\pi]-\left\{\frac{\pi}{2}\right\}}\]
is chosen as the principal branch.
This convention is adopted in NCERT, CBSE, JEE Main, JEE Advanced and most university-level mathematics.
Thus, the inverse secant function is defined as
\[\boxed{\sec^{-1}:(-\infty,-1]\cup[1,\infty)\rightarrow[0,\pi]-\left\{\frac{\pi}{2}\right\}}\]
📘 Definition
🔢 Fundamental Identities
🤔 Why is the Second Identity Restricted?
Consider
\[\sec^{-1}\left(\sec\frac{5\pi}{3}\right)\]
Since
\[\sec\frac{5\pi}{3}=2\]
therefore
\[\sec^{-1}(2)=\frac{\pi}{3}\]
not
\[\frac{5\pi}{3}\]
This is because the inverse secant function always returns the principal value belonging to
\[[0,\pi]-\left\{\frac{\pi}{2}\right\}\]
🗒️ Graphical Interpretation
On the interval
\[[0,\pi]-\left\{\frac{\pi}{2}\right\}\]
the secant function has two continuous branches separated by the vertical asymptote at
\[x=\frac{\pi}{2}\]
Each value belonging to
\[(-\infty,-1]\cup[1,\infty)\]
is attained exactly once. Therefore, the function satisfies the horizontal line test and possesses a unique inverse.
✏️ Example
Solved Example
1
Question
Evaluate \(\sec^{-1}(2)\)
  1. 1
    Recall the angle whose secant equals 2
  2. 2
    Check whether it lies in the principal interval.
\[\sec\frac{\pi}{3}=2\]
Since
\[\frac{\pi}{3}\in[0,\pi]-\left\{\frac{\pi}{2}\right\}\]
therefore
\[\boxed{\sec^{-1}(2)=\frac{\pi}{3}}\]
2
Question
Evaluate \(sec^{-1}(-2)\)
\[\sec\frac{2\pi}{3}=-2\]
Therefore,
\[\boxed{\sec^{-1}(-2)=\frac{2\pi}{3}}\]
3
Question
Evaluate \(\sec^{-1}\left(\sec\frac{4\pi}{3}\right)\)
\[\sec\frac{4\pi}{3}=-2\]
Now,
\[\sec^{-1}(-2)=\frac{2\pi}{3}\]
Hence,
\[\boxed{\sec^{-1}\left(\sec\frac{4\pi}{3}\right)=\frac{2\pi}{3}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\sec^{-1}x=\dfrac1{\sec x}\).
  • Confusing inverse secant with reciprocal secant.
  • Forgetting that \(\sec^{-1}x\) exists only for \(|x|\ge1\).
  • Including \(\dfrac{\pi}{2}\) in the principal interval.
  • Using \(\sec^{-1}(\sec x)=x\) for every real number.
📋 CBSE Case Study (HOTS)

An engineer measures the angle of elevation of a transmission tower. The measured secant value is

\[\sec\theta=2\]

The software immediately computes

\[\theta=\sec^{-1}(2)\]

Questions

  1. Why is the inverse secant function used?
  2. Why is the answer unique?
  3. Why is \(\dfrac{\pi}{2}\) excluded from the principal interval?
  4. What is the domain of the inverse secant function?
📐

Inverse Tangent Function (Principal Branch of Tangent Function)

📋
Table of Contents
13 sections
🗺️ Overview
Unlike the sine and cosine functions, the tangent function has range equal to the set of all real numbers. However, its domain excludes odd multiples of \(\dfrac{\pi}{2}\), where the function is undefined. Although its range is already suitable for defining an inverse function, the tangent function is still periodic and therefore not one-one on its entire domain. Hence, its domain must be restricted before defining the inverse tangent function.
📘 Domain and Range of the Tangent Function
📌 Restriction of Domain
🎨 SVG Diagram
Graphical Representation of \(\tan x\) with it Principal Branch
−3π/2 −π/2 π/2 3π/2 +1 −1 −π 0 π x y = tan(x) PRINCIPAL BRANCH: (−π/2, π/2)
📎 Side Note
Other Possible Restricted Domains
The tangent function is also one-one on every interval of length \(\pi\) between two consecutive vertical asymptotes. Some examples are
\[\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right),\quad\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\quad\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\]
Each of these intervals has the same range \(\mathbb{R}\)
Therefore, an inverse tangent function can be defined on each interval. Each such inverse is called a branch of the inverse tangent function.
📌 Principal Branch of the Inverse Tangent Function
🎨 SVG Diagram
Graphical Representation of \(\tan^{-1} x\)
Graph of arctan(x) principal branch with one additional periodic cycle above and below Deep sea blue themed plot showing the principal branch of y = tan^-1(x) highlighted in bright cyan between asymptotes y = -pi/2 and y = pi/2, with two additional branches y = arctan(x) + pi and y = arctan(x) - pi shown dimmer above and below, each with their own dashed asymptotes. π/2 -π/2 3π/2 -3π/2 0 -8 -6 -4 -2 2 4 6 8 principal branch: y = tan⁻¹(x) y = tan⁻¹(x) + π y = tan⁻¹(x) - π
📘 Definition of the Inverse Tangent Function
🔢 Fundamental Identities
🤔 Why is the Second Identity Restricted?
Consider
\[\tan^{-1}\left(\tan\frac{3\pi}{4}\right)\]
Since
\[\tan\frac{3\pi}{4}=-1\]
therefore,
\[\tan^{-1}(-1)=-\frac{\pi}{4},\]
not
\[\frac{3\pi}{4}\]
This is because the inverse tangent function always returns a value belonging to
\[\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\]
🗒️ Graphical Interpretation
On the interval
\[\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\]
the tangent function is continuous and strictly increasing from
\[-\infty\quad\text{to}\quad +\infty\]
Every horizontal line intersects the graph exactly once. Hence, the tangent function satisfies the horizontal line test and possesses a unique inverse.
✏️ Example
Solved Example
1
Question
Evaluate \(\tan^{-1}(1)\)
  1. 1
    Recall the standard angle whose tangent is 1.
  2. 2
    Verify that the angle belongs to the principal interval.
\[\tan\frac{\pi}{4}=1\]
Since
\[\frac{\pi}{4}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\]
therefore
\[\boxed{\tan^{-1}(1)=\frac{\pi}{4}}\]
2
Question
Evaluate \(\tan^{-1}(-\sqrt3)\)
\[\tan\left(-\frac{\pi}{3}\right)=-\sqrt3\]
Therefore,
\[\boxed{\tan^{-1}(-\sqrt3)=-\frac{\pi}{3}}\]
3
Question
Evaluate \(\tan^{-1}\left(\tan\frac{5\pi}{6}\right)\)
\[\tan\frac{5\pi}{6}=-\frac1{\sqrt3}\]
Now,
\[\tan^{-1}\left(-\frac1{\sqrt3}\right)=-\frac{\pi}{6}\]
Hence,
\[\boxed{\tan^{-1}\left(\tan\frac{5\pi}{6}\right)=-\frac{\pi}{6}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\tan^{-1}x=\dfrac1{\tan x}\).
  • Including the endpoints \(\pm\dfrac{\pi}{2}\) in the principal interval.
  • Using \(\tan^{-1}(\tan x)=x\) for every real number.
  • Ignoring the periodicity of the tangent function.
  • Confusing the domain of \(\tan x\) with the domain of \(\tan^{-1}x\).
📋 CBSE Case Study (HOTS)

An engineer determines the slope of a road and finds that

\[ \tan\theta=\frac34. \]

The surveying software computes

\[ \theta=\tan^{-1}\left(\frac34\right). \]

Questions

  1. Why is the inverse tangent function used instead of the tangent function?
  2. Why does the software return only one angle?
  3. What is the principal value interval of the inverse tangent function?
  4. Why are the endpoints excluded from the interval?
🗒️ 
📐

Inverse Cotangent Function (Principal Branch of Cotangent Function)

📋
Table of Contents
13 sections
📘 Definition
📌 Domain and Range of the Cotangent Function
🎨 SVG Diagram
Graphical Representation of \(\cot^{-1} x\)
Graph of cot^-1(x) principal branch, range zero to pi, with one additional periodic cycle above and below Deep sea blue themed plot showing the principal branch of y = cot^-1(x) = pi/2 minus arctan(x), continuous, highlighted in amber, range from 0 to pi, with two dimmer periodic branches above and below shifted by plus and minus pi. -π -π/2 0 π/2 π 3π/2 2π -10 -8 -6 -4 -2 2 4 6 8 10 principal branch: y = cot⁻¹(x), range (0, π) y = cot⁻¹(x) + π y = cot⁻¹(x) - π
🗒️ Restriction Of Domain
To define the inverse cotangent function uniquely, the domain of the cotangent function is restricted to the interval
\[\boxed{(0,\pi)}\]
The endpoints are excluded because
\[\cot0\quad\text{and}\quad\cot\pi\]
are undefined.

On the interval
\[(0,\pi),\]
the cotangent function is continuous and strictly decreasing from
\[+\infty\quad\text{to}\quad-\infty\]
Thus, every real number is attained exactly once. Hence,
\[\cot:(0,\pi)\rightarrow\mathbb{R}\]
is one-one and onto, and therefore invertible.
📎 Other Possible Restricted Domains
The cotangent function is one-one on every interval of length \(\pi\) lying between two consecutive multiples of \(\pi\). Examples include
\[(-\pi,0),\quad(0,\pi),\quad(\pi,2\pi)\]
Each of these intervals has the same range
\[\mathbb{R}\]
Therefore, an inverse cotangent function can be defined on each interval. Every such inverse is called a branch of the inverse cotangent function.
📌 Principal Branch of the Inverse Cotangent Function
📘 Definition of the Inverse Cotangent Function
🔢 Fundamental Identities
🤔 Did You Know?
Why is the Second Identity Restricted?
Consider
\[\cot^{-1}\left(\cot\frac{5\pi}{4}\right)\]
Since
\[\cot\frac{5\pi}{4}=1\]
therefore
\[\cot^{-1}(1)=\frac{\pi}{4}\]
not
\[\frac{5\pi}{4}\]
This happens because the inverse cotangent function always returns the principal value belonging to
\[(0,\pi)\]
🗒️ Graphical Interpretation
On the interval
\[(0,\pi)\]
the cotangent function is continuous and strictly decreasing from
\[+\infty\quad\text{to}\quad-\infty\]
Every horizontal line intersects the graph exactly once. Therefore, the cotangent function satisfies the horizontal line test and possesses a unique inverse.
✏️ Example
Solved Example
1
Question
Evaluate \(\cot^{-1}(1)\)
  1. 1
    Recall the angle whose cotangent equals 1
  2. 2
    Check whether it belongs to the principal interval.
\[\cot\frac{\pi}{4}=1\]
Since
\[\frac{\pi}{4}\in(0,\pi)\]
therefore
\[\boxed{\cot^{-1}(1)=\frac{\pi}{4}}\]
2
Question
Evaluate \(\cot^{-1}(-\sqrt3)\)
\[\cot\frac{5\pi}{6}=-\sqrt3\]
Since
\[\frac{5\pi}{6}\in(0,\pi)\]
therefore
\[\boxed{\cot^{-1}(-\sqrt3)=\frac{5\pi}{6}}\]
3
Question
Evaluate \(\cot^{-1}\left(\cot\frac{7\pi}{6}\right)\)
\[\cot\frac{7\pi}{6}=\sqrt3\]
Now,
\[\cot^{-1}(\sqrt3)=\frac{\pi}{6}\]
Hence,
\[\boxed{\cot^{-1}\left(\cot\frac{7\pi}{6}\right)=\frac{\pi}{6}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\cot^{-1}x=\dfrac1{\cot x}\).
  • Including the endpoints \(0\) and \(\pi\) in the principal interval.
  • Using \(\cot^{-1}(\cot x)=x\) for every real number.
  • Ignoring the periodicity of the cotangent function.
  • Confusing reciprocal functions with inverse functions.
📋 CBSE Case Study (HOTS)

An engineer measures the ratio of the adjacent side to the opposite side of a right triangle and finds

\[\cot\theta=\sqrt3\]

The calculator immediately evaluates

\[\theta=\cot^{-1}(\sqrt3)\]

Questions

  1. Why is the inverse cotangent function used?
  2. Why does the calculator return only one angle?
  3. Why is the interval \((0,\pi)\) selected as the principal branch?
  4. What is the domain of the inverse cotangent function?
📐

Properties of Inverse Trigonometric Functions

📋
Table of Contents
6 sections
🗺️ Overview
Inverse trigonometric functions possess several important algebraic and functional properties that are extensively used in NCERT, CBSE Board examinations, JEE Main, JEE Advanced, NDA, CUET and various engineering entrance examinations. These properties help simplify expressions, solve equations and prove trigonometric identities.
🗂️ Properties of Inverse Trigonometric Functions
  • 1. Fundamental Inverse Properties If a trigonometric function is restricted to its principal branch, then composing the function with its inverse gives the identity function.
    Inverse followed by Trigonometric Function
    \[\sin(\sin^{-1}x)=x,\quad -1\le x\le1\]
    \[\cos(\cos^{-1}x)=x,\quad -1\le x\le1\]
    \[\tan(\tan^{-1}x)=x,\quad x\in\mathbb{R}\]
    \[\cot(\cot^{-1}x)=x,\quad x\in\mathbb{R}\]
    \[\sec(\sec^{-1}x)=x,\quad |x|\ge1\]
    \[\operatorname{cosec}(\operatorname{cosec}^{-1}x)=x,\quad |x|\ge1\]
    These identities are always true because the inverse function exactly reverses the original function.
  • 2. Principal Value Properties When the inverse function acts after the trigonometric function, the identity holds only when the angle belongs to the principal value interval. \)
    Function Identity Principal Interval
    \(\sin^{-1}\) \(\sin^{-1}(\sin x)=x\) \(-\frac{\pi}{2}\le x\le\frac{\pi}{2}\)
    \(\cos^{-1}\) \(\cos^{-1}(\cos x)=x\) \(0\le x\le\pi\)
    \(\tan^{-1}\)\(\tan^{-1}(\tan x)=x\) \(-\frac{\pi}{2} < x < \frac{\pi}{2}\)
    \(\cot^{-1}\) \(\cot^{-1}(\cot x)=x\) \(0 < x < \pi\)
    \(\sec^{-1}\) \( \sec^{-1}(\sec x)=x\) \([0,\pi]-\left\{\frac{\pi}{2}\right\}\)
    \(\operatorname{cosec}^{-1}\) \( \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x\) \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}\)
    Outside these intervals, the answer must always be converted into the corresponding principal value.
  • 3. Principal Value Ranges
    Inverse Function Domain Range (Principal Value)
    \(\sin^{-1}x\) \([-1,1]\) \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \)
    \(\cos^{-1}x\) \([-1,1]\) \( [0,\pi] \)
    \(\tan^{-1}x\) \(\mathbb{R}\) \( \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \)
    \(\cot^{-1}x\) \(\mathbb{R}\) \( (0,\pi) \)
    \(\sec^{-1}x\) \( (-\infty,-1]\cup[1,\infty) \) \( [0,\pi]-\left\{\frac{\pi}{2}\right\} \)
    \(\operatorname{cosec}^{-1}x\) \( (-\infty,-1]\cup[1,\infty) \) \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\} \)
  • 4. Complementary Angle Properties These identities are among the most frequently used in Board examinations and competitive entrance tests.
    \[\boxed{\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}}\]
    for
    \[-1\le x\le1.\]
    \[\boxed{\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}} \]
    for all
    \[x\in\mathbb{R}\]
    \[\boxed{\sec^{-1}x+\operatorname{cosec}^{-1}x=\frac{\pi}{2}}\]
    for
    \[|x|\ge1\]
    Proof of the First Identity
    Let
    \[\theta=\sin^{-1}x\]
    Then
    \[\sin\theta=x\]
    Since
    \[\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta=x\]
    therefore
    \[\cos^{-1}x=\frac{\pi}{2}-\theta\]
    Hence,
    \[\boxed{\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}}\]
  • Properties for Negative Arguments Inverse trigonometric functions inherit symmetry from their corresponding trigonometric functions.
    \[\boxed{\sin^{-1}(-x)=-\sin^{-1}(x)}\]
    \[\boxed{\tan^{-1}(-x)=-\tan^{-1}(x)}\]
    \[\boxed{\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)}\]
    These functions are odd functions.

    The remaining inverse trigonometric functions satisfy
    \[\boxed{\cos^{-1}(-x)=\pi-\cos^{-1}(x)}\]
    \[\boxed{\cot^{-1}(-x)=\pi-\cot^{-1}(x)}\]
    \[\boxed{\sec^{-1}(-x)=\pi-\sec^{-1}(x)}\]
    These functions are neither odd nor even.
  • 6. Reciprocal Relations For permissible values of \(x\),
    \[\boxed{\sec^{-1}(x)=\cos^{-1}\left(\frac1x\right),\quad |x|\ge1}\]
    \[\boxed{\operatorname{cosec}^{-1}(x)=\sin^{-1}\left(\frac1x\right),\quad |x|\ge1}\]
    \[\boxed{\cot^{-1}(x)=\tan^{-1}\left(\frac1x\right),\quad x>0}\]
    For negative values,
    \[\boxed{\cot^{-1}(x)=\pi+\tan^{-1}\left(\frac1x\right),\quad x<0}\]
  • 7. Monotonic Nature
    Inverse Function Nature
    \(\sin^{-1}x\) Strictly Increasing
    \(\cos^{-1}x\) Strictly Decreasing
    \(\tan^{-1}x\) Strictly Increasing
    \(\cot^{-1}x\) Strictly Decreasing
    \(\sec^{-1}x\) Increasing on each branch
    \(\operatorname{cosec}^{-1}x\) Decreasing on each branch
  • 8. Useful Conversion Formulae
    \[\boxed{\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right),\quad |x|<1}\]
    \[\boxed{\cos^{-1}x=\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right),\quad 0
    \[\boxed{\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)}\]
    \[\boxed{\tan^{-1}x=\cos^{-1}\left(\frac1{\sqrt{1+x^2}}\right),\quad x\ge0}\]
    These identities are extensively used in integration and differentiation.
✏️ Example
Solved Example
1
Question
Evaluate \(\sin^{-1}\left(\frac35\right)+\cos^{-1}\left(\frac35\right)\)
Complementary angle property.
\[ \sin^{-1}\left(\frac35\right)+ \cos^{-1}\left(\frac35\right). \]
\[= \frac{\pi}{2}\]
2
Question
Evaluate \(\tan^{-1}(5)+\cot^{-1}(5)\)
\[\tan^{-1}(5)+\cot^{-1}(5)\]
\[=\frac{\pi}{2}\]
3
Question
SImplify \(\cos^{-1}(-x)\)
\[\boxed{\cos^{-1}(-x)=\pi-\cos^{-1}(x)}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\sin^{-1}x=\dfrac1{\sin x}\).
  • Using \(\sin^{-1}(\sin x)=x\) for every real number.
  • Ignoring principal value ranges.
  • Using reciprocal identities without checking domain restrictions.
  • Applying complementary angle identities outside their permissible domains.
🌟 Board and Competitive Exam Significance
  • Every CBSE Board examination includes questions based on principal values and inverse identities.
  • JEE Main and JEE Advanced frequently test complementary identities and conversion formulae.
  • These properties are extensively used in differentiation and integration.
  • Many trigonometric equations become simple using these identities.
  • Understanding these properties reduces calculation time and minimizes errors in objective examinations.
📐

Example 1

📋
Table of Contents
7 sections
❓ Question
Find the principal value of \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
💡 Concept
📖 Theory
🗺️ Roadmap
  1. Recall the standard angles whose sine is \(\dfrac{1}{\sqrt2}\).
  2. Select the angle lying in the principal interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\).
  3. Write the principal value.
🧩 Solution
Step-by-step Solution
  1. We know that
    \[\sin\frac{\pi}{4}=\frac{1}{\sqrt2}\]
  2. Also,
    \[\frac{\pi}{4}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\]
  3. Hence,
    \[\boxed{\sin^{-1}\left(\frac{1}{\sqrt2}\right)=\frac{\pi}{4}}\]
Verification
  1. \[\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt2},\]
    which confirms the result
🔑 Key Takeaway
  1. Whenever evaluating an inverse sine function, first identify the standard angle and then ensure that it lies in the principal value interval
    \[\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\]
⚡ Exam Tip
📐

Example 2

📋
Table of Contents
7 sections
❓ Question
Find the principal value of
\[ \cot^{-1}\left(\frac{1}{\sqrt3}\right). \]
💡 Concept
📖 Theory
🗺️ Roadmap
  1. Recall the standard angle whose cotangent is \(\dfrac{1}{\sqrt3}\).Verify that the angle lies in the principal interval \((0,\pi)\).Write the principal value.
🧩 Solution
Step-by-step Solution
  1. We know that
    \[\cot\frac{\pi}{3}=\frac{1}{\sqrt3}\]
  2. Also,
    \[\frac{\pi}{3}\in(0,\pi)\]
  3. Therefore,
    \[\boxed{\cot^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{3}}\]
Verification
  1. \[\cot\left(\frac{\pi}{3}\right)=\frac{1}{\sqrt3}\]
    which verifies the result
Alternative Angles
  1. Other angles such as
    \[ \frac{4\pi}{3}, \quad \frac{7\pi}{3}, \quad \frac{10\pi}{3}, \ldots \]
  2. also satisfy
    \[\cot\theta=\frac{1}{\sqrt3}\]
    but none of them belongs to the principal value interval \((0,\pi)\). Hence, they are not principal values.
🔑 Key Takeaway
  1. For the inverse cotangent function, always select the angle lying in the interval
    \[(0,\pi)\]
⚡ Exam Tip
📐

Example 3

📋
Table of Contents
6 sections
❓ Question
Show that

(i)

\[\sin^{-1}\left(2x\sqrt{1-x^2}\right)=2\sin^{-1}x,\quad -\frac{1}{\sqrt2}\le x\le\frac{1}{\sqrt2}\]

(ii)

\[\sin^{-1}\left(2x\sqrt{1-x^2}\right)=2\cos^{-1}x-\pi,\quad \frac{1}{\sqrt2}\le x\le1.\]
💡 Concept
🔢 Formula Used
🗺️ Roadmap
  1. Let the inverse trigonometric function be represented by an angle.
  2. Use the double-angle identity for sine.
  3. Determine the interval in which the resulting angle lies.
  4. Use the principal value property of the inverse sine function.
🧩 Solution
Solution (i)
  1. Let
    \[\theta=\sin^{-1}x\]
  2. Then
    \[\sin\theta=x\]
  3. where
    \[-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\]
  4. Since
    \[-\frac1{\sqrt2}\le x\le\frac1{\sqrt2}\]
  5. therefore,
    \[-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\]
  6. Hence,
    \[-\frac{\pi}{2}\le2\theta\le\frac{\pi}{2}\]
  7. Now,
    \[\begin{aligned} \cos\theta&=\sqrt{1-\sin^2\theta}\\&=\sqrt{1-x^2} \end{aligned}\]
  8. Using the double-angle identity
    \[\begin{aligned}\sin2\theta&=2\sin\theta\cos\theta\\&=2x\sqrt{1-x^2}\end{aligned}\]
  9. Therefore,
    \[\sin^{-1}\left(2x\sqrt{1-x^2}\right)=\sin^{-1}(\sin2\theta)\]
  10. Since
    \[2\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\]
  11. the principal value property gives
    \[\sin^{-1}(\sin2\theta)=2\theta\]
  12. Substituting
    \[\theta=\sin^{-1}x\]
  13. we obtain
    \[\boxed{\sin^{-1}\left(2x\sqrt{1-x^2}\right)=2\sin^{-1}x}\]
Solution (ii)
  1. Let
    \[\theta=\cos^{-1}x\]
  2. Then
    \[\cos\theta=x\]
  3. where
    \[0\le\theta\le\pi\]
  4. Since
    \[\frac1{\sqrt2}\le x\le1\]
  5. therefore,
    \[0\le\theta\le\frac{\pi}{4}\]
  6. Hence,
    \[0\le2\theta\le\frac{\pi}{2}\]
  7. Also,
    \[\begin{aligned}\sin\theta&=\sqrt{1-\cos^2\theta}\\&=\sqrt{1-x^2}\end{aligned}\]
  8. Using the double-angle identity,
    \[\begin{aligned}\sin2\theta&=2\sin\theta\cos\theta\\&=2x\sqrt{1-x^2}\end{aligned}\]
  9. Thus,
    \[\sin^{-1}\left(2x\sqrt{1-x^2}\right)=\sin^{-1}(\sin2\theta)\]
  10. Since
    \[0\le2\theta\le\frac{\pi}{2}\]
  11. we have
    \[\sin^{-1}(\sin2\theta)=2\theta\]
  12. Replacing
    \[\theta=\cos^{-1}x\]
  13. we obtain
    \[\boxed{\sin^{-1}\left(2x\sqrt{1-x^2}\right)=2\cos^{-1}x}\]
🔑 Key Takeaway
  1. Always verify that the angle lies within the principal value interval before removing the inverse trigonometric function.
  2. The double-angle identity alone is not sufficient; principal value restrictions must also be checked.
  3. Many JEE and CBSE questions are based on identifying the correct principal branch.
  4. Whenever an expression contains \(\sin^{-1}(\sin\theta)\), first determine the interval containing \(\theta\).
📐

Example 4

📋
Table of Contents
7 sections
❓ Question

Express

\[\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right), \quad -\frac{3\pi}{2} < x < \frac{\pi}{2}\]

in the simplest form.

🗒️ Concept Used
  • Trigonometric identities.
  • Principal value of the inverse tangent function.
  • Half-angle identity.
  • Sign analysis in different intervals.
🔢 Formula Used
🗺️ Roadmap
  1. Simplify the given trigonometric expression.
  2. Convert it into the tangent of an angle.
  3. Determine the interval of the angle.
  4. Use the principal value property of the inverse tangent function.
🧩 Solution
  1. Consider
    \[\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)\]
  2. Multiplying numerator and denominator by
    \[1+\sin x\]
  3. we obtain
    \[=\tan^{-1}\left(\frac{\cos x(1+\sin x)}{1-\sin^2x}\right)\]
  4. Using
    \[1-\sin^2x=\cos^2x\]
  5. we get
    \[=\tan^{-1}\left(\frac{1+\sin x}{\cos x}\right)\]
  6. Now apply the identity
    \[\frac{1+\sin x}{\cos x}=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\]
  7. Therefore,
    \[=\tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right]\]
  8. Let
    \[\theta=\frac{\pi}{4}+\frac{x}{2}\]
  9. Since
    \[-\frac{3\pi}{2} < x < \frac{\pi}{2}\]
  10. therefore,
    \[-\frac{\pi}{2} < \theta < \frac{\pi}{2}\]
  11. The principal value interval of the inverse tangent function is
    \[\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\]
  12. Hence,
    \[\tan^{-1}(\tan\theta)=\theta\]
  13. Substituting the value of \(\theta\),
    \[\boxed{\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\frac{\pi}{4}+\frac{x}{2}}\]
Alternative Method
  1. The identity
    \[\boxed{\frac{\cos x}{1-\sin x}=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}\]
  2. is a standard half-angle identity. Once this identity is recognized, the solution follows immediately after checking that
    \[\frac{\pi}{4}+\frac{x}{2}\]
    lies within the principal interval of the inverse tangent function.
🔑 Key Takeaway
  1. Always simplify trigonometric expressions before applying inverse functions.
  2. Half-angle identities are frequently used in inverse trigonometric problems.
  3. Never use \(\tan^{-1}(\tan\theta)=\theta\) without verifying that \(\theta\) belongs to the principal interval \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).
  4. Checking the interval is as important as simplifying the expression.
⚡ Exam Tip
📐

Example 5

📋
Table of Contents
7 sections
❓ Question
Write
\[\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right),\quad x > 1 \]
in the simplest form.
💡 Concept
🔢 Formula used
🗺️ Roadmap

  1. Assume the given inverse cotangent equals an angle.

  2. Construct a right triangle using the given cotangent ratio.

  3. Find the remaining trigonometric ratios.

  4. Identify the corresponding inverse trigonometric function.

  5. Write the simplest equivalent expression.

🧩 Solution
  1. Let
    \[\theta=\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)\]
  2. Then
    \[\cot\theta=\frac{1}{\sqrt{x^2-1}},\quad0<\theta<\pi\]
  3. Since
    \[x>1\]
    both quantities are positive. Hence, \(\theta\) lies in the first quadrant.
  4. Construct a right triangle taking
    Adjacent = 1 Opposite = √(x² − 1) Hypotenuse = x θ
    • Adjacent side \(=1\)
    • Opposite side \(=\sqrt{x^2-1}\)
  5. Using Pythagoras theorem,
    \[\text{Hypotenuse}=\sqrt{1+\left(x^2-1\right)}=x\]
  6. Therefore,
    \[\sec\theta=\frac{x}{1}=x\]
  7. Hence,
    \[\theta=\sec^{-1}x\]
  8. Since
    \[\theta=\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)\]
  9. therefore,
    \[\boxed{\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\sec^{-1}x,\quad x>1}\]
Alternative Method
  1. From the right triangle,
    \[\cos\theta=\frac1x\]
  2. Therefore,
    \[\theta=\cos^{-1}\left(\frac1x\right)\]
  3. Since
    \[\sec^{-1}x=\cos^{-1}\left(\frac1x\right),\quad x>1\]
  4. we again obtain
    \[\boxed{\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\sec^{-1}x}\]
🔑 Key Takeaway

  1. Whenever expressions involve \(\sqrt{x^2-1}\), constructing a right triangle is usually the quickest method.

  2. The condition \(x>1\) guarantees that the angle lies in the first quadrant.

  3. Remember the useful identity

    \[\boxed{\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\sec^{-1}x,\quad x>1}\]

⚡ Exam Tip
📐

Example 6

📋
Table of Contents
7 sections
❓ Question
Find the value of \(\sin^{-1}\left(\sin\frac{3\pi}{5}\right)\)
💡 Concept
📖 Theory
🗺️ Roadmap

  1. Check whether the given angle belongs to the principal value interval.

  2. Find an equivalent angle having the same sine value.

  3. Ensure that the equivalent angle lies in \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\).

  4. Write the principal value.

🧩 Solution
  1. We have
    \[\sin^{-1}\left(\sin\frac{3\pi}{5}\right)\]
  2. Since
    \[\frac{3\pi}{5} > \frac{\pi}{2}\]
    the angle does not belong to the principal value interval of the inverse sine function.
  3. Using the identity
    \[\sin(\pi-\theta)=\sin\theta\]
  4. we obtain
    \[\sin\frac{3\pi}{5}=\sin\left(\pi-\frac{3\pi}{5}\right)=\sin\frac{2\pi}{5}\]
  5. Now,
    \[\frac{2\pi}{5}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\]
  6. Therefore,
    \[\begin{aligned}\sin^{-1}\left(\sin\frac{3\pi}{5}\right)&=\sin^{-1}\left(\sin\frac{2\pi}{5}\right)\\&=\frac{2\pi}{5}\end{aligned}\]
  7. Hence,
    \[\boxed{\sin^{-1}\left(\sin\frac{3\pi}{5}\right)=\frac{2\pi}{5}}\]
Verification
  1. Since
    \[\sin\frac{3\pi}{5}=\sin108^\circ=\sin72^\circ=\sin\frac{2\pi}{5}\]
  2. and
    \[\frac{2\pi}{5}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\]
    the obtained answer is the correct principal value.
🔑 Key Takeaway
⚡ Exam Tip
· Updated

Academia Aeternum · Class XII · Mathematics

Chapter 2 — Inverse Trigonometric Functions

A self-contained interactive engine: concepts, formulas, a rule-based step-by-step solver, tips, common mistakes, original practice problems and hands-on exploration modules — no external API required.

NCERT · CBSE & Competitive Exams

01 Why "Inverse" Trig Functions Need a Restricted Range

Trigonometric functions like sin x are periodic and many-to-one — infinitely many values of x give the same sine. A function can only have an inverse if it is one-to-one and onto (bijective). So before we can "undo" sine, cosine or tangent, we chop their domain down to an interval where the function becomes strictly monotonic — this chosen interval is called the principal value branch.

Every inverse trig function is therefore really: "the unique angle, taken from one specific interval, whose sine/cosine/tangent equals a given number." Change the interval and you'd be talking about a different (but related) branch — that's the root of most sign-related confusion in this chapter.

02 Domain and Range (Principal Value Branches)

These six rows are the single most important table in the chapter — nearly every question checks whether you remember them exactly.

Function Domain Range (Principal Branch)
sin⁻¹x [−1, 1] [−π/2, π/2]
cos⁻¹x [−1, 1] [0, π]
tan⁻¹x ℝ (all reals) (−π/2, π/2)
cot⁻¹x ℝ (all reals) (0, π)
sec⁻¹x ℝ − (−1, 1) [0, π] − {π/2}
cosec⁻¹x ℝ − (−1, 1) [−π/2, π/2] − {0}

Notice the pattern: sin⁻¹, cosec⁻¹, tan⁻¹ live symmetrically around 0; cos⁻¹, cot⁻¹ live on [0, π]; and sec⁻¹, cosec⁻¹ exclude values strictly between −1 and 1 because secant and cosecant themselves never take those values.

03 Reading the Graphs

Every inverse trig graph is the mirror image of its parent function's restricted piece, reflected across the line y = x. That's why sin⁻¹x looks like a stretched "S" through the origin, while cos⁻¹x is a decreasing curve from (−1, π) down to (1, 0) — it never dips below the x-axis because its range starts at 0, not −π/2.

04 Property Group I — Reciprocal & Negative-Argument Identities

These simply restate a value using a "partner" inverse function, or flip the sign of the input.

sin⁻¹(1/x) = cosec⁻¹(x)   and   cos⁻¹(1/x) = sec⁻¹(x)    (|x| ≥ 1)
tan⁻¹(1/x) = cot⁻¹(x)    (x > 0)

sin⁻¹(−x) = −sin⁻¹(x)     tan⁻¹(−x) = −tan⁻¹(x)     cosec⁻¹(−x) = −cosec⁻¹(x)
cos⁻¹(−x) = π − cos⁻¹(x)     cot⁻¹(−x) = π − cot⁻¹(x)     sec⁻¹(−x) = π − sec⁻¹(x)

Odd-looking split? It's because sin⁻¹, tan⁻¹, cosec⁻¹ have ranges centred at 0 — so negating the input just negates the angle. But cos⁻¹, cot⁻¹, sec⁻¹ have ranges starting at 0 — so negating the input reflects the angle about π/2 instead, giving the "π − ..." form.

05 Property Group II — Complementary-Angle Identities

Just like sin θ = cos(π/2 − θ) in ordinary trigonometry, the inverse pairs are complementary too:

sin⁻¹x + cos⁻¹x = π/2    for x ∈ [−1, 1]
tan⁻¹x + cot⁻¹x = π/2    for x ∈ ℝ
sec⁻¹x + cosec⁻¹x = π/2    for |x| ≥ 1

These three identities alone can simplify a huge fraction of exam questions instantly — always check if an expression is secretly one of these three sums in disguise.

06 Property Group III — Sum & Difference of tan⁻¹

This is the most "algebra-heavy" and most tested identity family. It mirrors the tangent addition formula, but the branch (adding or subtracting π) depends on the sign of xy:

tan⁻¹x + tan⁻¹y = tan⁻¹( (x+y)/(1−xy) )     if xy < 1
tan⁻¹x + tan⁻¹y = π + tan⁻¹( (x+y)/(1−xy) )    if xy > 1, x > 0, y > 0
tan⁻¹x + tan⁻¹y = −π + tan⁻¹( (x+y)/(1−xy) )   if xy > 1, x < 0, y < 0

tan⁻¹x − tan⁻¹y = tan⁻¹( (x−y)/(1+xy) )     if xy > −1

The xy check exists because tan⁻¹x + tan⁻¹y can genuinely exceed the range (−π/2, π/2) that a single tan⁻¹ can output — so the formula must add or subtract π to "correct" the branch.

07 Property Group IV — Double-Angle Transformations

Substituting x = tan θ into the double-angle formulas of ordinary trigonometry converts 2 tan⁻¹x into three equivalent inverse forms — extremely useful for simplification and equation-solving:

2 tan⁻¹x = sin⁻¹( 2x / (1+x²) )    for |x| ≤ 1
2 tan⁻¹x = cos⁻¹( (1−x²) / (1+x²) )    for x ≥ 0
2 tan⁻¹x = tan⁻¹( 2x / (1−x²) )    for −1 < x < 1

The trick behind all three: let θ = tan⁻¹x, so x = tan θ, then apply sin 2θ, cos 2θ, tan 2θ in terms of tan θ and rewrite the result back as an inverse function of 2θ.

08 Principal Value — What "Evaluate" Actually Means

When a question says "find the principal value of sin⁻¹(1/2)", it is asking for exactly one number: the unique angle inside sin⁻¹'s range, [−π/2, π/2], whose sine is 1/2. That number is π/6 — not 5π/6, even though sin(5π/6) is also 1/2, because 5π/6 lies outside the principal branch. This single idea — "pick the angle from the allowed range, not just any angle with the right ratio" — is the conceptual core of this entire chapter.

🧮 Rule-Based Step-by-Step Solver

Pick a problem type below. Every step is generated locally in your browser using the exact NCERT identities — there is no external AI API call involved.

Σ Domain & Range

sin⁻¹x : [−1,1] → [−π/2, π/2]
cos⁻¹x : [−1,1] → [0, π]
tan⁻¹x : ℝ → (−π/2, π/2)
cot⁻¹x : ℝ → (0, π)
sec⁻¹x : |x|≥1 → [0,π]−{π/2}
cosec⁻¹x : |x|≥1 → [−π/2,π/2]−{0}

Σ Reciprocal Identities

sin⁻¹(1/x) = cosec⁻¹x, |x|≥1
cos⁻¹(1/x) = sec⁻¹x, |x|≥1
tan⁻¹(1/x) = cot⁻¹x, x>0
tan⁻¹(1/x) = −π+cot⁻¹x, x<0

Σ Negative-Argument Identities

sin⁻¹(−x) = −sin⁻¹x
tan⁻¹(−x) = −tan⁻¹x
cosec⁻¹(−x) = −cosec⁻¹x
cos⁻¹(−x) = π−cos⁻¹x
cot⁻¹(−x) = π−cot⁻¹x
sec⁻¹(−x) = π−sec⁻¹x

Σ Complementary Identities

sin⁻¹x + cos⁻¹x = π/2, x∈[−1,1]
tan⁻¹x + cot⁻¹x = π/2, x∈ℝ
sec⁻¹x + cosec⁻¹x = π/2, |x|≥1

Σ Sum & Difference (tan⁻¹)

tan⁻¹x+tan⁻¹y = tan⁻¹((x+y)/(1−xy)), xy<1
  = π+tan⁻¹((x+y)/(1−xy)), xy>1,x,y>0
  = −π+tan⁻¹((x+y)/(1−xy)), xy>1,x,y<0
tan⁻¹x−tan⁻¹y = tan⁻¹((x−y)/(1+xy)), xy>−1

Σ Double-Angle Forms of 2tan⁻¹x

2tan⁻¹x = sin⁻¹(2x/(1+x²)), |x|≤1
2tan⁻¹x = cos⁻¹((1−x²)/(1+x²)), x≥0
2tan⁻¹x = tan⁻¹(2x/(1−x²)), −1<x<1

Σ Standard Value Quick-Reference

x sin⁻¹x cos⁻¹x tan⁻¹x
0 0 π/2 0
1/2 π/6 π/3 —
1/√2 π/4 π/4 —
√3/2 π/3 π/6 —
1 π/2 0 π/4
1/√3 — — π/6
√3 — — π/3

💡 Ticks & Tips

🎯

Always name the range first

Before evaluating anything, write down the range of the function you're inverting. It filters out the wrong-branch answer before you even compute.

🔄

Substitute x = tan θ / x = sin θ to "see" the identity

Whenever an expression looks tangled (like 2tan⁻¹x or sin⁻¹(2x√(1−x²))), let x = tan θ or x = sin θ, rewrite fully in θ using double-angle formulas, simplify, then convert back. This turns algebra into pure trig.

➕

Watch the sign of xy in tan⁻¹ sum formulas

If xy > 1 you must add or subtract π — this is the single most common mark lost in board exams on this chapter.

🪞

cos⁻¹, cot⁻¹, sec⁻¹ "mirror about π/2" for negatives

Unlike sin⁻¹/tan⁻¹/cosec⁻¹ (which simply flip sign), these three give π minus the positive-value answer — memorise this as a pair, not six separate rules.

📐

Convert everything to one function before comparing

If a question mixes sin⁻¹ and cos⁻¹, use the complementary identity to write both in terms of the same function before equating or simplifying.

🧭

Degrees vs radians — commit to one, and check the domain matches

NCERT answers are almost always expected in radians. If you compute in degrees along the way, convert the final answer back before writing it down.

✅

After simplifying, sanity-check with a rough numeric value

Plug in an approximate decimal for x and check both sides of your final identity roughly match on a calculator — this catches sign errors in seconds.

🧩

sec⁻¹ and cosec⁻¹ questions: check |x| ≥ 1 first

A huge number of errors come from trying to evaluate sec⁻¹ or cosec⁻¹ at a value strictly between −1 and 1 — such an expression is simply undefined, not "small".

⚠️ Common Mistakes

✗

Writing sin⁻¹(sin x) = x for every x

Wrong: assuming this holds for all real x.

Correct: sin⁻¹(sin x) = x only when x ∈ [−π/2, π/2]. Outside this interval you must first reduce x to an equivalent angle inside that range.

✗

Treating sin⁻¹x as 1/(sin x)

Wrong: reading the "−1" as a reciprocal power.

Correct: sin⁻¹x denotes the inverse function ("arcsine"), completely different from (sin x)⁻¹ = 1/sin x = cosec x.

✗

Forgetting the ± π correction in tan⁻¹x + tan⁻¹y

Wrong: always using tan⁻¹((x+y)/(1−xy)) blindly.

Correct: check the sign of xy first — if xy > 1, the true sum needs +π (both positive) or −π (both negative) added to that expression.

✗

Applying cos⁻¹(−x) = −cos⁻¹x

Wrong: assuming cos⁻¹ behaves like an odd function.

Correct: cos⁻¹(−x) = π − cos⁻¹x, since cos⁻¹'s range [0, π] is not symmetric about zero.

✗

Evaluating sec⁻¹(0.5) or cosec⁻¹(0.5)

Wrong: plugging in a value between −1 and 1.

Correct: sec⁻¹ and cosec⁻¹ are undefined for |x| < 1 — always verify the domain |x| ≥ 1 before attempting to evaluate.

✗

Picking the "obvious" angle instead of the principal-branch angle

Wrong: sin⁻¹(1/2) = 5π/6 (because sin(5π/6) = 1/2 too).

Correct: sin⁻¹(1/2) = π/6, since only π/6 lies inside sin⁻¹'s range [−π/2, π/2].

✗

Mixing degree and radian answers mid-solution

Wrong: writing tan⁻¹(1) = 45 in an equation full of π's.

Correct: keep the whole solution in radians (tan⁻¹1 = π/4) unless the question explicitly asks for degrees.

✗

Assuming 2 sin⁻¹x = sin⁻¹(2x)

Wrong: pulling constants straight through the inverse function.

Correct: inverse trig functions are not linear — 2 sin⁻¹x has no such simple single-function form; only 2 tan⁻¹x has the neat double-angle conversions shown in Property Group IV.

✏️ Concept-Building Practice (Original Problems)

Organised by the property group it strengthens. Click a question to reveal the full worked solution.

🕹️ Interactive Modules

Choose a function, then drag the slider to see how the input x, the principal-value output angle, and the graph point all move together — live, inside the real domain/range boundaries.

x
x = 0.00

Drag x and watch both sides of a complementary identity update live — instant numerical proof that the identity holds across the whole domain.

x
x = 0.00
—
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Master NCERT Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions with these comprehensive, exam-oriented notes designed for CBSE Board Exams, JEE Main, JEE Advanced, CUET, NDA and other competitive examinations. This chapter explains the concept of inverse trigonometric functions, principal value branches, domain and range restrictions, properties, identities, standard values, and simplification techniques through detailed theory, step-by-step derivations and fully solved examples.…
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    Frequently Asked Questions

    Inverse trigonometric functions return the principal value of an angle corresponding to a given trigonometric ratio. The six inverse functions are \(\sin^{-1}x\), \(\cos^{-1}x\), \(\tan^{-1}x\), \(\cot^{-1}x\), \(\sec^{-1}x\), and \(\csc^{-1}x\).

    Trigonometric functions are periodic and not one-one over their entire domains. Restricting the domain to a principal interval makes them one-one, allowing inverse functions to exist.

    The principal value range of \(\sin^{-1}x\) is \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), while its domain is \([-1,1]\).

    \(\sin^{-1}x:\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), \(\cos^{-1}x:[0,\pi]\), \(\tan^{-1}x:\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), \(\cot^{-1}x:(0,\pi)\), \(\sec^{-1}x:[0,\pi]-\left\{\dfrac{\pi}{2}\right\}\), and \(\csc^{-1}x:\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]-\{0\}\).

    The domain of \(\tan^{-1}x\) is the set of all real numbers \(\mathbb{R}\), and its principal value lies in \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).

    First determine whether \(x\) lies in the principal interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). If not, reduce it to an equivalent angle within the principal interval before applying the inverse sine function.

    For every \(x\in[-1,1]\), the identity is \(\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\). It is one of the most frequently used identities in Board and JEE examinations.

    Common mistakes include confusing inverse functions with reciprocal functions, ignoring principal value intervals, using \(\sin^{-1}(\sin x)=x\) for all real \(x\), and forgetting domain restrictions.

    They are extensively used in differentiation, integration, limits, trigonometric equations, coordinate geometry and calculus. Questions based on principal values and identities are regularly asked in CBSE Board and competitive examinations.

    Memorize the principal value intervals, learn standard inverse values, practice principal value problems from different quadrants, revise important identities, and always check the domain before simplifying inverse trigonometric expressions.

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