# 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(π/6) = sin(−π/6)

Range of the principal value of sn

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

^{-1}(-1/2) is -π/6.^{-1}(√3/2)

**Answer**

Let cos

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

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

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

^{-1}(√3/2) is π/6.^{-1}(2)

**Answer**

Let cosec

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

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

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

^{-1}(2) is π/6.^{-1}(√3)

**Answer**

Let tan

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

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

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

Therefore, the principal value of tan

^{-1}(-√3) is -π/3^{-1}(-1/2)

**Answer**

Let cos

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

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

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

^{-1}(-1/2) is 2π^{-1}(-1)

**Answer**

Let tan

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

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

^{-1}(−1) is -π/4.^{-1}(2/√3)

**Answer**

Let sec

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

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

^{-1}(2/√3) is π/6.^{-1}(√3)

**Answer**

Let cot

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

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

^{-1}√3 is π.^{-1}(-1/√2)

**Answer**

Let cos

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

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

^{-1}is [0, π] and cos (3π4) = -1/√2.
Therefore, the principal value of cos

^{-1}(-1/√2) is 3π/4.
10. cosec

^{-1}(-√2)**Answer**

Let cosec

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

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

^{-1}(-√2) is -π/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(π/4)

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

^{-1}is (−π/2, π/2).
∴ tan

^{-1}(1) = π/4
Let cos

^{-1}(−1/2) = y,
then cos y = −1/2 = −cosπ/3 = cos (π − π/3)

= cos (2π/3)

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

^{-1}is [0, π].
∴ cos

^{-1}(−1/2) = 2π/3
Let sin

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

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

^{-1}is [-/π2, π/2].
∴ sin

^{-1}(-1/2) = -π/6
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

12. cos

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

^{-1}(1/2)

**Answer**

Let cos

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

cos x = 1/2 = cos π/3

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

∴ cos

^{-1}(1/2)

= π/3

Let sin

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

sin y = 1/2

= sin π/6

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

^{-1}is [-π/2, π/2].

∴ sin

^{-1}(1/2) = π/6

Now,

cos

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

^{-1}(1/2)

= π/3 + 2×π/6

= π/3 + π/3

= 2π/3

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

**Answer**

It is given that sin

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

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

Hence, the option (B) is correct.

14. tan

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

(B) -π/3

(C) π/3

(D) 2π/3

**Answer**

Let tan

^{-1}√3 = x,then
tan x = √3 = tan π/3

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

^{-1}is (-π/2, π/2).
∴ tan

^{-1}√3 = π/3
Let sec

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

= sec (π - π/3)

= sec (2π/3)

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

^{-1}is [0, π]- {π/2}
∴ sec

^{-1}(-2) =2π/3
Now,

tan

^{-1}√3 - sec^{-1}(-2)
= π/3 - 2π/3

= -π/3

Hence, the option (B) is correct.