Inverse Trigonometric Functions

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Class 12 Maths Chapter 2
Inverse Trigonometric Function
Important Questions

Preparing for the Class 12 Maths exams, specifically Chapter 2 on Inverse Trigonometric Functions, holds paramount importance in building a robust mathematical foundation. This chapter elucidates the inverse relationships of trigonometric functions, unraveling the methodologies to solve equations involving inverse trigonometric functions. To excel in this chapter, familiarizing oneself with essential questions becomes pivotal. Here are some class 12 Inverse Trigonometric Function important questions and answers. These questions are crafted to evaluate comprehension of the core concepts within this segment. Interacting with a diverse array of question formats aids in navigating uncertainties, ensuring comprehensive readiness for the impending exams.

Table of Contents

Introduction

The concept of inverse trigonometric functions is an important area in mathematics, particularly in calculus and geometry. These functions are the inverses of the trigonometric functions and are used to find the angle when the value of the trigonometric function is known. Let’s delve into an introduction to inverse trigonometric functions that the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse, cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse,tangent of an angle is the ratio of the sine to the cosine of that angle, or equivalently, the ratio of the opposite side to the adjacent side. Questions on inverse trigonometric functions will help students in understanding the concepts better.

What are Inverse Trigonometric Function?

Inverse trigonometric functions are functions that reverse the operation of the basic trigonometric functions—sine, cosine, and tangent. When you know the ratio of the sides of a right triangle (as in trigonometric functions), you use trigonometric functions to find the angles. Inverse trigonometric functions, on the other hand, allow you to determine the angle when you know the ratio of the sides.

Class 12 Inverse Trigonometric Function Important Questions and Answers

Q1. The principal value of

$\\[2.5 bp]\tan^{-1}\left(\tan \dfrac{3π}{5}\right)$ is:

Options

(a) $\dfrac{2π}{5}\\[4.5 bp]$ (b) $\dfrac{-2π}{5}\\[4.5 bp]$ (c) $\dfrac{3π}{5}\\[4.5 bp]$ (d) $\dfrac{-3π}{5}\\[4.5 bp]$

Ans.(b)
Explanation:
$\tan^{-1} \left(\tan \dfrac{3π}{5}\right)\\$ This can be written as:
$\tan^{-1} \left(\tan \dfrac{3π}{5}\right) = \tan^{-1} \left(\tan\left[π – \dfrac{2π}{5}\right]\right)\\[4.5 bp] = \tan^{-1} \left(- \tan \dfrac{2π}{5}\right) \{\text{since }\tan(π – x) = -\tan x\}\\[4.5 bp] = –\tan^{-1} \left(\tan \dfrac{2π}{2}\right)\\[4.5 bp] = –\dfrac{2π}{5}\\[4.5 bp]$

Q2.

$\sin\left[\dfrac{π}{3} – \sin^{-1}\left(-\dfrac{1}{2}\right)\right]$ is equal to:

Options

(a) $\dfrac{1}{2}\\[4.5 bp]$ (b) $\dfrac{1}{3}\\[4.5 bp]$ (c) -1
(d) 1

Ans. (d)

Explanation:
$\sin\left[\dfrac{π}{3} – \sin^{-1}\left(-\dfrac{1}{2}\right)\right]\\[4.5 bp] = \sin\left[\dfrac{π}{3} – \sin^{-1}\left[\sin \left(-\dfrac{π}{6}\right)\right]\right]\\[4.5 bp] \sin\left[\dfrac{π}{3} – \left(-\dfrac{π}{6} \right)\right]\\[4.5 bp] = \sin\left(\dfrac{π}{3} + \dfrac{π}{6}\right)\\[4.5 bp] = \sin \dfrac{π}{2}\\[4.5 bp] = 1$

Q3. Write the value of

$\\[4.5 bp ]\tan^{-1} (\sqrt{3}) – \cot^{-1} (- \sqrt{3})$

Explanation:
$\text{We have,} \tan^{-1} (\sqrt{3}) – \cot^{-1} (-\sqrt{3})\\[4.5 bp] = \tan^{-1} (\sqrt{3}) – \{π – \cot^{-1} (\sqrt{3})\} \\[4.5 bp] [∵ \cot^{-1} (- x) = π – \cot^{-1} x; x ∈ R]\\[4.5 bp] = \tan^{-1}\sqrt{3} – π + \cot^{-1}\sqrt{3}\\[4.5 bp] = (\tan^{-1} \sqrt{3} + \cot^{-1} \sqrt{3}) – π\\[4.5 bp] = -\dfrac{π}{2}\\[4.5 bp]$

Q4. Write the principal value of

$\\[2.5 bp]cos^{-1} [\cos(680)^o]$

Explanation:
First we check the given angle lies in the principal value branch. If it is so, then use the property $\cos^{-1} (\cos θ) = θ, ∀θ ∈ [0, 180°].$ Otherwise reduce the angle such that it lies in the principal value branch.

Q5. Find the value of

$\cos^{-1} \left(\dfrac{1}{2}\right) + 2 \sin^{-1} \left(\dfrac{1}{2}\right)$

Explanation:
Let us take, $\text{y = cos}^{-1} \left(\dfrac{1}{2} \right) \\$ This can be written as:
$\\ \cos\space y = \left( \dfrac{1}{2} \right)\\[4.5 bp] \cos \space y = \cos \left( \dfrac{\pi }{3}\right).\\$ Thus, the range of principal value of $\cos^{-1} \text{is [0, }\pi]\\$ Therefore, the principal value of
$\cos^{-1} \left(\dfrac{1}{2}\right) \space \text{is } \dfrac{\pi}{3}.$

CBSE Class 12 Maths Chapter wise Important Questions

Chapter No.	Chapter Name
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Applications of Derivatives
Chapter 7	Integrals
Chapter 8	Applications of the Integrals
Chapter 9	Differential Equations
Chapter 10	Vectors
Chapter 11	Three - dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

Conclusion

The concepts of relations and functions are integral to the understanding of mathematical principles and are widely applied in various fields of study. Exploring CBSE class 12 maths inverse trignonometric function important questions and answers, becomes essential for a comprehensive grasp of this fundamental aspect of mathematics. To enhance understanding, platforms like oswal.io provide a wealth of resources, including question-answer sets, comprehensive class 12 questions and answers.Utilising these resources ensures students acquire the knowledge necessary to excel in their studies and examinations, inviting them to delve deeper into the captivating realm of inverse trignonometric function.

Frequently Asked Questions

Q1 : What Are Inverse Trigonometric Functions?

Ans: Inverse trigonometric functions are the inverses of the standard trigonometric functions (sine, cosine, tangent, etc.). They are used to find the angle when the value of a trigonometric function is known.

Q2 : Why Are Inverse Trigonometric Functions Important?

Ans: They are important for solving equations involving trigonometric functions and are widely used in calculus, particularly in integration and differentiation. They also have practical applications in physics, engineering, and other sciences.

Q3 : How Do You Define the Inverse Sine Function?

Ans: The inverse sine function, written as sin⁻¹(x) or arcsin(x), gives the angle whose sine is x. Its domain is [-1, 1], and its range is

\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]

Q4 : What is the Difference Between sin⁻¹(x) and 1/sin(x)?

Ans: Sin⁻¹(x) is the inverse sine function, giving the angle for a given sine value, while

\dfrac{1}{\sin(x)}

is the cosecant function, which is the reciprocal of the sine function.

Q5: Can Inverse Trigonometric Functions Be Defined for All Real Numbers?

Ans: No, inverse trigonometric functions have specific domain restrictions. For example, arcsin(x) and arccos(x) are only defined for values between -1 and 1, while arctan(x) and arccot(x) can accept any real number as input.

Chapter Wise Important Questions for CBSE Board Class 12 Maths
Relations and Functions
Inverse Trigonometric Functions
Matrices
Determinants
Continuity and Differentiability
Applications of Derivatives
Integrals
Applications of the Integrals
Differential Equations
Vectors
Three - dimensional Geometry
Linear Programming
Probability

Class 12 Maths Chapter 2 Inverse Trigonometric Function Important Questions

Introduction

What are Inverse Trigonometric Function?

Class 12 Inverse Trigonometric Function Important Questions and Answers

Q1. The principal value of

\\[2.5 bp]\tan^{-1}\left(\tan \dfrac{3π}{5}\right) is:

Options

(a) \dfrac{2π}{5}\\[4.5 bp] (b) \dfrac{-2π}{5}\\[4.5 bp] (c) \dfrac{3π}{5}\\[4.5 bp] (d) \dfrac{-3π}{5}\\[4.5 bp]

Q2.

\sin\left[\dfrac{π}{3} – \sin^{-1}\left(-\dfrac{1}{2}\right)\right] is equal to:

Options

(a) \dfrac{1}{2}\\[4.5 bp] (b) \dfrac{1}{3}\\[4.5 bp] (c) -1 (d) 1

Q3. Write the value of

\\[4.5 bp ]\tan^{-1} (\sqrt{3}) – \cot^{-1} (- \sqrt{3})

Q4. Write the principal value of

\\[2.5 bp]cos^{-1} [\cos(680)^o]

Q5. Find the value of

\cos^{-1} \left(\dfrac{1}{2}\right) + 2 \sin^{-1} \left(\dfrac{1}{2}\right)

CBSE Class 12 Maths Chapter wise Important Questions

Conclusion

Frequently Asked Questions

Q1 : What Are Inverse Trigonometric Functions?

Q2 : Why Are Inverse Trigonometric Functions Important?

Q3 : How Do You Define the Inverse Sine Function?

Q4 : What is the Difference Between sin⁻¹(x) and 1/sin(x)?

Q5: Can Inverse Trigonometric Functions Be Defined for All Real Numbers?

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Class 12 Maths Chapter 2
Inverse Trigonometric Function
Important Questions

$\\[2.5 bp]\tan^{-1}\left(\tan \dfrac{3π}{5}\right)$ is:

(a) $\dfrac{2π}{5}\\[4.5 bp]$ (b) $\dfrac{-2π}{5}\\[4.5 bp]$ (c) $\dfrac{3π}{5}\\[4.5 bp]$ (d) $\dfrac{-3π}{5}\\[4.5 bp]$

$\sin\left[\dfrac{π}{3} – \sin^{-1}\left(-\dfrac{1}{2}\right)\right]$ is equal to:

(a) $\dfrac{1}{2}\\[4.5 bp]$ (b) $\dfrac{1}{3}\\[4.5 bp]$ (c) -1
(d) 1

$\\[4.5 bp ]\tan^{-1} (\sqrt{3}) – \cot^{-1} (- \sqrt{3})$

$\\[2.5 bp]cos^{-1} [\cos(680)^o]$

$\cos^{-1} \left(\dfrac{1}{2}\right) + 2 \sin^{-1} \left(\dfrac{1}{2}\right)$