# NCERT Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-2 (Inverse Trigonometric Functions)Exercise 2.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Class 12 Mathematics

Chapter-2 (Inverse Trigonometric Functions)

Questions and answers given in practice

Chapter-2 (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

= (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

= (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.