Class 12 NCERT Mathematics Solutions : Chapter 2 Inverse Trigonometric Functions - Exercise 2.1

In this article, we will be going to solve the entire exercise 2.1 of our NCERT textbook which is Inverse Trigonometric Functions. Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. It's fundamental in many areas of mathematics and applied sciences, including physics, engineering, and astronomy.

Inverse Trigonometric Functions

Inverse Trigonometric Functions are functions that reverse the action of the basic trigonometric functions (sine, cosine, tangent, etc.). They allow us to find the angle when the value of the trigonometric function is known. Understanding these functions is crucial for solving equations and modelling situations where the angle needs to be found from a given trigonometric value.

Knowing How to Interpret Inverse Trigonometric Functions

The idea of inverse trigonometric functions—the opposites of common trigonometric functions like sine, cosine, tangent, etc.—is first presented to students in Exercise 2.1. The definition of these functions, their domains and ranges, and the process of determining principal values for particular inputs are the main points of emphasis in this section.

Qualities of Trigonometric Functions in Reverse

Understanding the fundamental characteristics of inverse trigonometric functions—including periodicity, symmetry, and behavior under composition with trigonometric functions—is essential for this exercise. Comprehending these attributes is crucial for resolving issues related to modifying and simplifying expressions that incorporate inverse trigonometric functions.

Methodical Solutions for Exercise 2.1 Issues

This section ensures that students understand the process needed to solve problems involving inverse trigonometric functions by offering comprehensive, step-by-step solutions to every problem in Exercise 2.1. The solutions deal with identifying principal values, reducing the complexity of expressions, and using identities to resolve equations.

Class 12 NCERT Mathematics Solutions: Inverse Trigonometric Functions - Exercise 2.1

Find the principal values of the following:

Question 1. sin^-1(-1/2)

Solution:

Let sin^-1(-1/2) = y then, sin y = -1/2

Range of principal value for sin^-1 is [-π/2,π/2] and sin(-π/6)=-1/2.

Therefore, principal value of sin^-1(-1/2)=-π/6.

Question 2. cos^-1(√3/2)

Solution:

Let cos^-1(√3/2) = y then, cos y = √3/2

Range of principal value for cos^-1 is [0, π] and cos(π/6) = √3/2

Therefore, principal value of cos^-1(√3/2) = π/6.

Question 3. cosec^-1(2)

Solution:

Let cosec^-1(2) = y then, cosec y = 2

Range of principal value for cosec^-1 is [-π/2, π/2] -{0}  and cosec(π/6) = 2

Therefore, principal value of cosec^-1(2) = π/6.

Question 4: tan^-1(-√3)

Solution:

Let tan^-1(-√3) = y then, tan y = -√3

Range of principal value for tan^-1 is (-π/2, π/2) and tan(-π/3) = -√3

Therefore, principal value of tan^-1(-√3) = -π/3.

Question 5. cos^-1(-1/2)

Solution:

Let cos^-1(-1/2) = y then, cos y = -1/2

Range of principal value for cos^-1 is [0, π] and cos(2π/3) = -1/2

Therefore, principal value of cos^-1(-1/2) = 2π/3.

Question 6. tan-1(-1)

Solution:

Let tan^-1(-1) = y then, tan y = -1

Range of principal value for tan^-1 is (-π/2, π/2)  and tan(-π/4) = -1

Therefore, principal value of tan^-1(-1) = -π/4.

Question 7. sec^-1(2/√3)

Solution:

Let sec^-1(2/√3) = y then, sec y = 2/√3

Range of principal value for sec^-1 is [0, π] - {π/2} and sec(π/6) = 2/√3

Therefore, principal value of sec^-1(2/√3) = π/6.

Question 8. cot^-1(√3)

Solution:

Let cot^-1(√3) = y then, cot y = √3

Range of principal value for cot^-1 is (0, π) and cot(π/6) = √3

Therefore, principal value of cot^-1(√3) = π/6.

Question 9. cos^-1(-1/√2)

Solution:

Let cos^-1(-1/√2) = y then, cos y = -1/√2

Range of principal value for cos^-1 is [0, π] and cos(2π/3) = -1/2

Therefore, principal value of cos^-1(-1/2) = 3π/4.

Question 10. cosec^-1(-√2)

Solution:

Let cosec^-1(-√2) = y then, cosec y = -√2

Range of principal value for cosec^-1 is [-π/2, π/2] -{0} and cosec(-π/4) = -√2

Therefore, principal value of cosec^-1(-√2) = -π/4.

Find the values of the following:

Question 11. tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)

Solution:

For solving this question we will use principal values of sin^-1, cos^-1 & tan^-1

Let sin^-1(-1/2) = y then, sin y = -1/2

Range of principal value for sin^-1 is [-π/2, π/2] and sin(-π/6) = -1/2.

Therefore, principal value of sin^-1(-1/2) = -π/6.

Let cos^-1(-1/2) = x then, cos x = -1/2

Range of principal value for cos^-1 is [0, π] and cos(2π/3) = -1/2

Therefore, principal value of cos^-1(-1/2) = 2π/3.

Let tan^-1(1) = z then, tan z = -1

Range of principal value for tan^-1 is (-π/2, π/2)  and tan(π/4) = 1

Therefore, principal value of tan^-1(1) = π/4.

Now, tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2) = π/4 + 2π/3 - π/6 

Adding them we will get, 

= (3π + 8π - 2π)/12 

= 9π/12

= 3π/4 

Question 12. cos^-1(1/2) + 2 sin^-1(1/2)

Solution:

For solving this question we will use principal values of sin^-1 & cos^-1

Let sin^-1(1/2) = y then, sin y = -1/2

Range of principal value for sin^-1 is [-π/2, π/2] and sin(π/6) = 1/2.

Therefore, principal value of sin^-1(1/2) = π/6.

Let cos^-1(1/2) = x then, cos x = 1/2

Range of principal value for cos^-1 is [0, π] and cos(π/3) = 1/2

Therefore, principal value of cos^-1(1/2) = π/3.

Now, cos^-1(1/2) + 2 sin^-1(1/2) = π/3 + 2π/6 

Adding them we will get,

= (2π + 2π)/6

= 4π/6

= 2π/3 

Question 13. If sin^–1 x = y, then

(A) 0 ≤ y ≤ π (B) -π / 2 ≤y ≤ π / 2 (C) 0 < y < π (D) -π / 2 <y < π / 2

Solution:

We know that the principal range for sin-1 is [-π / 2, π / 2]

Hence, if sin-1 x = y, y € [-π / 2, π / 2]

Therefore, -π / 2 ≤y ≤ π / 2.

Hence, option (B) is correct.

Question 14. tan^–1(√3) - sec^-1(-2) is equal to

(A) π (B) -π/3 (C) π/3 (D) 2π/3 

Solution:

For solving this question we will use principal values of sec^-1 & tan^-1

Let tan^-1(√3) = y then, tan y = √3

Range of principal value for tan^-1 is (-π/2, π/2) and tan(π/3) = √3

Therefore, principal value of tan^-1(√3) = π/3.

Let sec^-1(-2) = y then, sec y = -2

Range of principal value for sec^-1 is [0, π] - {π/2} and sec(2π/3) = - 2

Therefore, principal value of sec^-1(-2) = 2π/3.

Now, tan^–1 (√3) - sec ^-1(-2) 

= π/3 - 2π/3

= -π/3

Hence, option (B) is correct.

Related Articles,

• Trigonometric Functions
• Trigonometric Functions of Sum and Difference of Two Angles
• Inverse trigonometric Functions

Conclusion

The purpose of Exercise 2.1 in the Class 12 NCERT Mathematics textbook is to introduce the inverse trigonometric functions and highlight their principal values and domains and their properties. Students will obtain a basic understanding of manipulating and applying inverse trigonometric functions in a variety of mathematical contexts by completing this exercise.