# NCERT Solutions for Class 12th: Ch 2 Inverse Trigonometric Functions Exercise 2.1

#### NCERT Solutions for Class 12th: Ch 2 Inverse Trigonometric Functions Exercise 2.1 Math

Page No: 41

**Exercise 2.1**

Find the principal values of the following:

1. sin

1. sin

^{-1}(-1/2)**Answer**

1. Let sin

^{-1}(−1/2) = y, then
sin y = −1/2 = −sin(Ï€/6) = sin(−Ï€/6)

Range of the principal value of sn

^{-1}is [-Ï€/2, Ï€/2] and sin -Ï€/6) = -1/2
Therefore, the principal value of sin

^{-1}(-1/2) is -Ï€/6.^{-1}(√3/2)

**Answer**

Let cos

^{-1}(√3/2) = y,
cos y = √3/2 = cos (Ï€/6)

We know that the range of the principal value branch of cos

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

^{-1}(√3/2) is Ï€/6.^{-1}(2)

**Answer**

Let cosec

^{-1}(2) = y.
Then, cosec y = 2 = cosec (Ï€/6)

We know that the range of the principal value branch of cosec

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

^{-1}(2) is Ï€/6.^{-1}(√3)

**Answer**

Let tan

^{-1}(-√3) = y,
then tan y = -√3 = -tan Ï€/3 = tan (-Ï€/3)

We know that the range of the principal value branch of tan

^{-1}is (-Ï€/2, Ï€/2) and tan (-Ï€/3)
= -√3

Therefore, the principal value of tan

^{-1}(-√3) is -Ï€/3^{-1}(-1/2)

**Answer**

Let cos

^{-1}(-1/2) = y,
then cos y = -1/2 = -cos Ï€/3 = cos (Ï€-Ï€/3) = cos (2Ï€/3)

We know that the range of the principal value branch of cos

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

^{-1}(-1/2) is 2Ï€^{-1}(-1)

**Answer**

Let tan

^{-1}(-1) = y. Then, tan y = -1 = -tan (Ï€/4) = tan (-Ï€/4)
We know that the range of the principal value branch of tan

^{-1}is (-Ï€/2, Ï€/2) and tan (-Ï€/4) = -1.
Therefore, the principal value of tan

^{-1}(−1) is -Ï€/4.^{-1}(2/√3)

**Answer**

Let sec

^{-1}(2/√3) = y, then sec y = 2/√3 = sec (Ï€/6)
We know that the range of the principal value branch of sec

^{-1}is [0, Ï€] − {Ï€/2} and sec (Ï€/6) = 2/√3.
Therefore, the principal value of sec

^{-1}(2/√3) is Ï€/6.^{-1}(√3)

**Answer**

Let cot

^{-1}√3 = y, then cot y = √3 = cot (Ï€/6).
We know that the range of the principal value branch of cot

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

^{-1}√3 is Ï€.^{-1}(-1/√2)

**Answer**

Let cos

^{-1}(-1/√2) = y,
then cos y = -1/√2 = -cos (Ï€/4) = cos (Ï€ - Ï€/4) = cos (3Ï€/4).

We know that the range of the principal value branch of cos

^{-1}is [0, Ï€] and cos (3Ï€4) = -1/√2.
Therefore, the principal value of cos

^{-1}(-1/√2) is 3Ï€/4.
10. cosec

^{-1}(-√2)**Answer**

Let cosec

^{-1}(−√2) = y, then cosec y = −√2 = −cosec (Ï€/4) = cosec (−Ï€/4)
We know that the range of the principal value branch of cosec

^{-1}is [-Ï€/2, Ï€/2]-{0} and cosec(-Ï€/4) = -√2.
Therefore, the principal value of cosecc

^{-1}(-√2) is -Ï€/4.
Page No. 42

Find the values of the following:

11. tan

^{-1}(1) + cos^{-1}(-1/2) + sin^{-1}(-1/2)**Answer**

Let tan

^{-1}(1) = x,
then tan x = 1 = tan(Ï€/4)

We know that the range of the principal value branch of tan

^{-1}is (−Ï€/2, Ï€/2).
∴ tan

^{-1}(1) = Ï€/4
Let cos

^{-1}(−1/2) = y,
then cos y = −1/2 = −cosÏ€/3 = cos (Ï€ − Ï€/3)

= cos (2Ï€/3)

We know that the range of the principal value branch of cos

^{-1}is [0, Ï€].
∴ cos

^{-1}(−1/2) = 2Ï€/3
Let sin

^{-1}(−1/2) = z,
then sin z = −1/2 = −sin Ï€/6 = sin (−Ï€/6)

We know that the range of the principal value branch of sin

^{-1}is [-/Ï€2, Ï€/2].
∴ sin

^{-1}(-1/2) = -Ï€/6
Now,

tan

^{-1}(1) + cos^{-1}(-1/2) + sin^{-1}(-1/2)
= Ï€/4 + 2Ï€/3 − Ï€/6

= (3Ï€ + 8Ï€ − 2Ï€)/12

= 9Ï€/12 = 3Ï€/4

12. cos

^{-1}(1/2) + 2 sin

^{-1}(1/2)

**Answer**

Let cos

^{-1}(1/2) = x, then

cos x = 1/2 = cos Ï€/3

We know that the range of the principal value branch of cos−1 is [0, Ï€].

∴ cos

^{-1}(1/2)

= Ï€/3

Let sin

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

sin y = 1/2

= sin Ï€/6

We know that the range of the principal value branch of sin

^{-1}is [-Ï€/2, Ï€/2].

∴ sin

^{-1}(1/2) = Ï€/6

Now,

cos

^{-1}(1/2) + 2sin

^{-1}(1/2)

= Ï€/3 + 2×Ï€/6

= Ï€/3 + Ï€/3

= 2Ï€/3

13. If sin

^{-1}x = y, then
(A) 0 ≤ y ≤ Ï€

(B) -Ï€/2 ≤ y ≤ Ï€/2

(C) 0 < y < Ï€

(D) -Ï€/2 < y < Ï€/2

**Answer**

It is given that sin

^{-1}x = y.
We know that the range of the principal value branch of sin

^{-1}is [-Ï€/2, Ï€/2].
Therefore, -Ï€/2 ≤ y ≤ Ï€/2.

Hence, the option (B) is correct.

14. tan

^{-1}√3 - sec^{-1}(-2) is equal to
(A) Ï€

(B) -Ï€/3

(C) Ï€/3

(D) 2Ï€/3

**Answer**

Let tan

^{-1}√3 = x,then
tan x = √3 = tan Ï€/3

We know that the range of the principal value branch of tan

^{-1}is (-Ï€/2, Ï€/2).
∴ tan

^{-1}√3 = Ï€/3
Let sec

^{-1}(-2) = y, then
sec y = -2 = -sec Ï€/3

= sec (Ï€ - Ï€/3)

= sec (2Ï€/3)

We know that the range of the principal value branch of sec

^{-1}is [0, Ï€]- {Ï€/2}
∴ sec

^{-1}(-2) =2Ï€/3
Now,

tan

^{-1}√3 - sec^{-1}(-2)
= Ï€/3 - 2Ï€/3

= -Ï€/3

Hence, the option (B) is correct.