#### 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}(âˆ’1/2) = y, then
sin y = âˆ’1/2 = âˆ’sin(Ï€/6) = sin(âˆ’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}(âˆš3/2)

**Answer**

Let cos

^{-1}(âˆš3/2) = y,
cos y = âˆš3/2 = cos (Ï€/6)

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

^{-1}Â is [0, Ï€] and cos (Ï€/6) = âˆš3/2
Therefore, the principal value of cos

^{-1}(âˆš3/2) 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}Â is [-Ï€/2, Ï€/2] - {0} and cosec (Ï€/6) = 2.
Therefore, the principal value of cosec

^{-1}(2) is Ï€/6.^{-1}(âˆš3)

**Answer**

Let tan

^{-1}(-âˆš3) = y,
then tan y = -âˆš3 = -tan Ï€/3 = tan (-Ï€/3)

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

^{-1}Â is (-Ï€/2, Ï€/2) and tan (-Ï€/3)
= -âˆš3

Therefore, the principal value of tan

^{-1}Â (-âˆš3) 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}Â 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}Â is (-Ï€/2, Ï€/2) and tan (-Ï€/4) = -1.
Therefore, the principal value of tan

^{-1}(âˆ’1) is -Ï€/4.^{-1}(2/âˆš3)

**Answer**

Let sec

^{-1}(2/âˆš3) = y, then sec y = 2/âˆš3 = sec (Ï€/6)
We know that the range of the principal value branch of sec

^{-1}Â is [0, Ï€] âˆ’ {Ï€/2} and sec (Ï€/6) = 2/âˆš3.
Therefore, the principal value of sec

^{-1}(2/âˆš3) is Ï€/6.^{-1}(âˆš3)

**Answer**

Let cot

^{-1}âˆš3 = y, then cot y = âˆš3 = cot (Ï€/6).
We know that the range of the principal value branch of cot

^{-1}Â is (0, Ï€) and cot (Ï€/6) = âˆš3.
Therefore, the principal value of cot

^{-1}âˆš3 is Ï€.^{-1}(-1/âˆš2)

**Answer**

Let cos

^{-1}(-1/âˆš2) = y,
then cos y = -1/âˆš2 = -cos (Ï€/4) = cos (Ï€ - Ï€/4) = cos (3Ï€/4).

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

^{-1}Â is [0, Ï€] and cos (3Ï€4) = -1/âˆš2.
Therefore, the principal value of cos

^{-1}(-1/âˆš2) is 3Ï€/4.
10. cosec

^{-1}(-âˆš2)Â Â**Answer**

Let cosec

^{-1}(âˆ’âˆš2) = y, then cosec y = âˆ’âˆš2 = âˆ’cosec (Ï€/4) = cosec (âˆ’Ï€/4)
We know that the range of the principal value branch of cosec

^{-1}Â is [-Ï€/2, Ï€/2]-{0} and cosec(-Ï€/4) = -âˆš2.
Therefore, the principal value of cosecc

^{-1}(-âˆš2) 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}Â is (âˆ’Ï€/2, Ï€/2).
âˆ´ tan

^{-1}(1) = Ï€/4
Let cos

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

= cos (2Ï€/3)

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

^{-1}Â is [0, Ï€].
âˆ´ cos

^{-1}(âˆ’1/2) = 2Ï€/3
Let sin

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

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

^{-1}Â is [-/Ï€2, Ï€/2].
âˆ´ sin

^{-1}(-1/2) = -Ï€/6
Now,

tan

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

= (3Ï€ + 8Ï€ âˆ’ 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 cosâˆ’1 is [0, Ï€].

âˆ´ 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}Â is [-Ï€/2, Ï€/2].

âˆ´ sin

^{-1}(1/2) = Ï€/6

Now,

cos

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

^{-1}(1/2)

= Ï€/3 + 2Ã—Ï€/6

= Ï€/3 + Ï€/3

= 2Ï€/3

13. If sin

^{-1}Â x = y, then
(A) 0 â‰¤ y â‰¤ Ï€

(B) -Ï€/2 â‰¤ y â‰¤ Ï€/2

(C) 0 < y < Ï€Â

(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}Â is [-Ï€/2, Ï€/2].
Therefore, -Ï€/2 â‰¤ y â‰¤ Ï€/2.

Hence, the option (B) is correct.

14. tan

^{-1}âˆš3 - sec^{-1}(-2) is equal to
(A) Ï€

(B) -Ï€/3

(C) Ï€/3

(D) 2Ï€/3

**Answer**

Let tan

^{-1}âˆš3 = x,then
tan x = âˆš3 = tan Ï€/3

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

^{-1}Â is (-Ï€/2, Ï€/2).
âˆ´ tan

^{-1}âˆš3 = 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}Â is [0, Ï€]- {Ï€/2}
âˆ´ sec

^{-1}(-2) =2Ï€/3
Now,

tan

^{-1}âˆš3 - sec^{-1}(-2)
= Ï€/3 - 2Ï€/3

= -Ï€/3

Hence, the option (B) is correct.