#### 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(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.

2. cos-1(√3/2)

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.

3. cosec-1(2)

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.

4. tan-1(√3)

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

5. cos-1(-1/2)

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

6. tan-1(-1)

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.

7. sec-1(2/√3)

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.

8. cot-1(√3)

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.

9. cos-1(-1/√2)

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)

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)

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)

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

It is given that sin-1x = 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

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.

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