CBSE Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

This resource provides detailed NCERT Solutions for Chapter 2 of CBSE Class 12 Mathematics, focusing on Inverse Trigonometric Functions. It covers finding the principal values for various inverse trigonometric functions like sine, cosine, cosecant, and tangent, including those with negative arguments. The solutions explain the concept of the principal value range for each function and demonstrate how to determine the correct principal value. This guide is essential for students preparing for their board examinations, offering clear explanations and step-by-step problem-solving techniques to build a strong foundation in this topic. Mastering these solutions will aid in exam revision and improve conceptual understanding.

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BoardCBSE
ClassClass 12
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 2

Chapter summary

Chapter 2, Inverse Trigonometric Functions, deals with the principal values of inverse trigonometric functions. This NCERT Solutions set focuses on Exercise 2.1, guiding students to find the principal values for functions such as $\sin^{-1}$, $\cos^{-1}$, $\csc^{-1}$, and $\tan^{-1}$. It emphasizes understanding the specific range of principal values for each function to correctly solve problems involving positive and negative arguments.

Learning outcomes

  • Understand the concept of principal values for inverse trigonometric functions.
  • Determine the principal value of $\sin^{-1}(x)$ for given values of x.
  • Determine the principal value of $\cos^{-1}(x)$ for given values of x.
  • Determine the principal value of $\csc^{-1}(x)$ for given values of x.
  • Determine the principal value of $\tan^{-1}(x)$ for given values of x.
  • Apply the range of principal values to solve inverse trigonometric problems.

Topics covered

Paper topics

  • Inverse Trigonometric Functions
  • Principal Value
  • Range of Inverse Trigonometric Functions
  • $\sin^{-1}(x)$
  • $\cos^{-1}(x)$
  • $\tan^{-1}(x)$
  • $\csc^{-1}(x)$
  • Solving Inverse Trigonometric Problems

Important topics

  • Principal Value Range of Inverse Trigonometric Functions
  • Finding Principal Values for Negative Arguments
  • Application of Principal Value Ranges

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Questions and Solutions

Question 1

Find the principal value of \sin^{-1}\left(-\frac{1}{2}\right).
Solution:

Let the principal value of \sin^{-1}\left(-\frac{1}{2}\right) be y.

By definition of the inverse sine function, we have \sin y = -\frac{1}{2}.

We know that \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}. Therefore, \sin y = -\sin\left(\frac{\pi}{6}\right).

Using the property -\sin(\theta) = \sin(-\theta), we get \sin y = \sin\left(-\frac{\pi}{6}\right).

The range of the principal value of \sin^{-1}(x) is \[-\frac{\pi}{2}, \frac{\pi}{2}\].

Since -\frac{\pi}{6} lies within this range, the principal value of \sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}.

Question 2

Find the principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right).
Solution:

Let the principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) be y.

By definition of the inverse cosine function, we have \cos y = \frac{\sqrt{3}}{2}.

We know that \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.

The range of the principal value of \cos^{-1}(x) is \[0, \pi\].

Since \frac{\pi}{6} lies within this range, the principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) is \frac{\pi}{6}.

Question 3

Find the principal value of \csc^{-1}(2).
Solution:

Let the principal value of \csc^{-1}(2) be y.

By definition of the inverse cosecant function, we have \csc y = 2.

We know that \csc\left(\frac{\pi}{6}\right) = 2.

The range of the principal value of \csc^{-1}(x) is \[-\frac{\pi}{2}, \frac{\pi}{2}\] - \{0\}.

Since \frac{\pi}{6} lies within this range, the principal value of \csc^{-1}(2) is \frac{\pi}{6}.

Question 4

Find the principal value of \tan^{-1}(-\sqrt{3}).
Solution:

Let the principal value of \tan^{-1}\left(-\sqrt{3}\right) be y.

By definition of the inverse tangent function, we have \tan y = -\sqrt{3}.

We know that \tan\left(\frac{\pi}{3}\right) = \sqrt{3}. Therefore, \tan y = -\tan\left(\frac{\pi}{3}\right).

Using the property -\tan(\theta) = \tan(-\theta), we get \tan y = \tan\left(-\frac{\pi}{3}\right).

The range of the principal value of \tan^{-1}(x) is \[-\frac{\pi}{2}, \frac{\pi}{2}\].

Since -\frac{\pi}{3} lies within this range, the principal value of \tan^{-1}\left(-\sqrt{3}\right) is -\frac{\pi}{3}.

Question 5

Find the principal value of \cos^{-1}\left(-\frac{1}{2}\right).
Solution:

Let the principal value of \cos^{-1}\left(-\frac{1}{2}\right) be y.

By definition of the inverse cosine function, we have \cos y = -\frac{1}{2}.

We know that \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}. Therefore, \cos y = -\cos\left(\frac{\pi}{3}\right).

Using the property -\cos(\theta) = \cos(\pi - \theta), we get \cos y = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right).

The range of the principal value of \cos^{-1}(x) is \[0, \pi\].

Since \frac{2\pi}{3} lies within this range, the principal value of \cos^{-1}\left(-\frac{1}{2}\right) is \frac{2\pi}{3}.

Common mistakes

  • Incorrectly recalling or applying the principal value ranges for different inverse trigonometric functions.
  • Errors in determining the correct angle for negative trigonometric values.
  • Confusing the principal value range with the general solution.

Revision tips

  • Memorize the principal value ranges for all six inverse trigonometric functions.
  • Practice identifying the correct quadrant for negative arguments.
  • Work through each example and exercise problem step-by-step to reinforce the method.
  • Pay close attention to the signs of the arguments and their impact on the principal value.

Practice MCQs

Q1. What is the principal value of $\sin^{-1}(-1/2)$?

Q2. The principal value of $\cos^{-1}(\sqrt{3}/2)$ is:

Q3. What is the principal value of $\csc^{-1}(2)$?

Q4. Find the principal value of $\tan^{-1}(-\sqrt{3})$:

Q5. Which range represents the principal values for $\sin^{-1}(x)$?

Frequently asked questions

What are NCERT Solutions for Class 12 Maths Chapter 2?

These are step-by-step solutions for the exercises in Chapter 2: Inverse Trigonometric Functions of the NCERT Class 12 Mathematics textbook, designed to help students understand and solve problems related to principal values.

What is the main focus of Chapter 2 NCERT Solutions?

The primary focus is on understanding and calculating the principal values of various inverse trigonometric functions like $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, and $\csc^{-1}$, adhering to their defined ranges.

Why is the principal value important in inverse trigonometric functions?

The principal value ensures that each input to an inverse trigonometric function corresponds to a unique output, making the function well-defined and single-valued.

What is the principal value range for $\sin^{-1}(x)$?

The principal value range for $\sin^{-1}(x)$ is $[-\pi/2, \pi/2]$.

What is the principal value range for $\cos^{-1}(x)$?

The principal value range for $\cos^{-1}(x)$ is $[0, \pi]$.

How do these solutions help in exam preparation?

These solutions provide clear, step-by-step methods to solve problems, helping students revise concepts, practice techniques, and build confidence for their board exams.

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