# Class 12 NCERT Solutions – Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.1

**Find the principal values of the following:**

**Question 1. sin**^{-1}(-1/2)

^{-1}(-1/2)

**Solution:**

Let sin

^{-1}(-1/2) = y then, sin y = -1/2Range 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)

^{-1}(√3/2)

**Solution:**

Let cos

^{-1}(√3/2) = y then, cos y = √3/2Range of principal value for cos

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

^{-1}(√3/2) = π/6.

**Question 3. cosec**^{-1}(2)

^{-1}(2)

**Solution:**

Let cosec

^{-1}(2) = y then, cosec y = 2Range of principal value for cosec

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

^{-1}(2) = π/6.

**Question 4: tan**^{-1}(-√3)

^{-1}(-√3)

**Solution:**

Let tan

^{-1}(-√3) = y then, tan y = -√3Range of principal value for tan

^{-1}is (-π/2, π/2) and tan(-π/3) = -√3Therefore, principal value of tan

^{-1}(-√3) = -π/3.

**Question 5. cos**^{-1}(-1/2)

^{-1}(-1/2)

**Solution:**

Let cos

^{-1}(-1/2) = y then, cos y = -1/2Range of principal value for cos

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

^{-1}(-1/2) = 2π/3.

**Question 6. tan-1(-1)**

**Solution:**

Let tan

^{-1}(-1) = y then, tan y = -1Range of principal value for tan

^{-1}is (-π/2, π/2) and tan(-π/4) = -1Therefore, principal value of tan

^{-1}(-1) = -π/4.

**Question 7. sec**^{-1}(2/√3)

^{-1}(2/√3)

**Solution:**

Let sec

^{-1}(2/√3) = y then, sec y = 2/√3Range of principal value for sec

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

^{-1}(2/√3) = π/6.

**Question 8. cot**^{-1}(√3)

^{-1}(√3)

**Solution:**

Let cot

^{-1}(√3) = y then, cot y = √3Range of principal value for cot

^{-1}is (0, π) and cot(π/6) = √3Therefore, principal value of cot

^{-1}(√3) = π/6.

**Question 9. cos**^{-1}(-1/√2)

^{-1}(-1/√2)

**Solution:**

Let cos

^{-1}(-1/√2) = y then, cos y = -1/√2Range of principal value for cos

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

^{-1}(-1/2) = 3π/4.

**Question 10. cosec**^{-1}(-√2)

^{-1}(-√2)

**Solution:**

Let cosec

^{-1}(-√2) = y then, cosec y = -√2Range of principal value for cosec

^{-1}is [-π/2, π/2] -{0} and cosec(-π/4) = -√2Therefore, 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)

^{-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 = -1/2Range 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 = -1/2Range of principal value for cos

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

^{-1}(-1/2) = 2π/3.Let tan

^{-1}(1) = z then, tan z = -1Range of principal value for tan

^{-1}is (-π/2, π/2) and tan(π/4) = 1Therefore, 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/6Adding them we will get,

= (3π + 8π – 2π)/12

= 9π/12

= 3π/4

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

^{-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 = -1/2Range 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 = 1/2Range of principal value for cos

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

^{-1}(1/2) = π/3.Now, cos

^{-1}(1/2) + 2 sin^{-1}(1/2) = π/3 + 2π/6Adding them we will get,

= (2π + 2π)/6

= 4π/6

= 2π/3

**Question 13. If sin**^{–1} x = y, then

^{–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

^{–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 = √3Range of principal value for tan

^{-1}is (-π/2, π/2) and tan(π/3) = √3Therefore, principal value of tan

^{-1}(√3) = π/3.Let sec

^{-1}(-2) = y then, sec y = -2Range of principal value for sec

^{-1}is [0, π] – {π/2} and sec(2π/3) = – 2Therefore, 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.