#### 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(-1/2)

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.

2. cos-1(√3/2)

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.

3. cosec-1(2)

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.

4. tan-1(√3)

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

5. cos-1(-1/2)

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€

6. tan-1(-1)

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.

7. sec-1(2/√3)

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.

8. cot-1(√3)

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€.

9. cos-1(-1/√2)

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)

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)

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)

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

It is given that sin-1x = 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

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.

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