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Class 12 NCERT Solutions – Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.1
• Last Updated : 25 Feb, 2021

### 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.

### 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

### (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.

### (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.

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