Inverse Trigonometry Functions(Ex. 2.1)

Exercise 2.1

Find the Principal values of the following:

1. sin^-1(-1/2)

Sol:- Let y = sin^-1(-1/2)………………..(1)

We know sin^-1(-x) = -sin^-1x

⇒ y = - sin^-1(1/2)

⇒ -y = sin^-1 (1/2)

⇒ sin (-y) = 1/2

⇒ sin (-y) = sin (Π/6)

⇒ -y = Π /6

⇒ y = - Π/6

From Eq. 1:- put the value of y

⇒ sin^-1(-1/2) = - Π/6

Ans… - Π/6 lies in the range of sin^-1x [-Π/2, Π/2],

           so principal value is - Π/6.

2. cos^-1(√3/2)

Sol:- Let y = cos^-1(√3/2) ………………..(1)

⇒ cos y = √3/2

⇒ cos y = cos (Π/6)

⇒ y = Π /6

From Eq. 1:- put the value of y

⇒ cos^-1(√3/2) = Π/6

Ans… Π/6 lies in the range of cos^-1x [0, Π],

           so principal value is Π/6.

3. cosec^-1(2)

Sol:- Let y = cosec^-1(2) ………………..(1)

⇒ cosec y = 2

⇒ cosec y = cosec (Π /6)

⇒ y = Π /6

From Eq. 1:- put the value of y

⇒ cosec^-1(2) = Π /6

Ans… Π/6 lies in the range of cosec^-1x  [-Π/2, Π/2]-{0},

           so principal value is Π/6.

4. tan^-1(-√3)

Sol:- y = tan^-1(-√3) ………………..(1)

We know tan^-1(-x) = - tan^-1x

⇒ y = - tan^-1(√3)

⇒ -y = tan^-1(√3)

⇒ tan (-y) = √3

⇒ tan (-y) = tan (Π/3)

⇒ tan (-y) = tan (Π/3)

⇒ -y = Π/3

⇒ y = -Π/3

From Eq. 1:- put the value of y

⇒ tan^-1(-√3) = - Π/3

Ans… - Π/3 lies in the range of tan^-1x  (-Π/2, Π/2),

           so principal value is - Π/3.

5. cos^-1(-1/2)

⇒ y = cos^-1(-1/2)………………..(1)

We know cos^-1(-x) = Π - cos^-1x

⇒ y = Π - cos^-1(1/2)

⇒ cos^-1(1/2) = Π - y

⇒ cos (Π -y) = 1/2

⇒ cos (Π -y) = cos (Π/3)

⇒ Π -y = Π/3

⇒ y = Π-Π/3

⇒ y = 2Π/3

From Eq. 1:- put the value of y

⇒ cos^-1(-1/2) = 2Π/3

Ans… 2Π/3 lies in the range of cos^-1x [0, Π],

           so principal value is 2Π/3.

6. tan^-1(-1)

Sol:- Let y = tan^-1(-1) ………………..(1)

We know tan^-1(-x) = -tan^-1x

⇒ y = -tan^-1(1)

⇒ -y = tan^-1(1)

⇒ tan (-y) = 1

⇒ tan (-y) = tan (Π/4)

⇒ -y = Π/4

⇒ y = - Π/4

From Eq. 1:- put the value of y

⇒ tan^-1(-1) = - Π/4

Ans… - Π/4 lies in the range of tan^-1x (-Π/2, Π/2),

           so principal value is - Π/4.

7. sec^-1(2/√3)

Sol:- y = sec^-1(2/√3) ………………..(1)

⇒ sec y = 2/√3

⇒ sec y = sec(Π/6)

⇒ y = Π/6

From Eq. 1:- put the value of y

⇒ sec^-1(2/√3) = Π/6

Ans… Π/6 lies in the range of sec^-1x [0,Π]-{Π/2},

           so principal value is Π/6.

8. cot^-1(√3)

Sol:- y = cot^-1(√3) ………………..(1)

⇒ cot y = √3

⇒ cot y = cot (Π/6)

⇒ y = Π/6

From Eq. 1:- put the value of y

⇒ cot^-1(√3) = Π/6

Ans… Π/6 lies in the range of cot^-1x (0, Π),

           so principal value is Π/6.

9. cos^-1(-1/√2)

Sol:- Let y = cos^-1(-1/√2) ………………..(1)

We know cos^-1(-x) = Π - cos^-1x

⇒ y = Π - cos^-1(1/√2)

⇒ cos^-1(1/√2) = Π - y

⇒ cos (Π-y) = 1/√2

⇒ cos (Π-y) = cos (Π/4)

⇒ Π - y = Π/4

⇒ y = Π - Π/4

⇒ y = 3Π/4

From Eq. 1:- put the value of y

⇒ cos^-1(-1/√2) = 3Π/4

Ans… 3Π/4 lies in the range of cos^-1x [0, Π],

           so principal value is 3Π/4.

10. cosec^-1(-√2)

Sol:- Let y = cosec^-1(-√2) ………………..(1)

We know cosec^-1(-x) = -cosec^-1x

⇒ y = -cosec^-1(√2)

⇒ -y = cosec^-1(√2)

⇒ cosec (-y) = √2

⇒ cosec (-y) = cosec (Π/4)

⇒ -y = Π/4

⇒ y = -Π/4

From Eq. 1:- put the value of y

⇒ cosec (-√2) = - Π/4

Ans… - Π/4 lies in the range of cosec^-1x [-Π/2, Π/2]-{0},

           so principal value is - Π/4.

Find the values of the following:

11. tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)

Sol:- tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2) ………………..(1)
A = tan^-1(1) ………………..(2)

⇒ tan (A) = 1

⇒ tan (A) = tan (Π/4)

⇒ A = Π/4 { lies in range of tan^-1x (-Π/2,Π/2) }

B = cos^-1(-1/2) ………………..(3)

We know cos^-1(-x) = Π- cos^-1x

⇒ B = Π - cos^-1(1/2)

⇒ cos^-1(1/2) = Π - B

⇒ cos (Π-B) = 1/2

⇒ cos (Π-B) = cos (Π/3)

⇒ Π - B = Π/3

⇒ B = Π - Π/3

⇒ B = 2Π/3 { lies in range of cos^-1x [0,Π] }

C = sin^-1(-1/2) ………………..(4)

We know sin^-1(-x) = -sin^-1x

⇒ C = -sin^-1(1/2)

⇒ -C = sin^-1(1/2)

⇒ sin(-C) = 1/2

⇒ sin(-C) = sin (Π/6)

⇒ -C = Π/6

⇒ C = - Π/6{ lies in range of sin^-1x [-Π/2,Π/2] }

put Eq. 2,3,4 values in Eq. 1:-

= A + B +C

put A, B, C values:-

= Π/4 + 2Π/3 + (-Π/6)

= (3Π+8Π-2Π)/12

= 9Π/12

= 3Π/4 (Ans……)

12. cos^-1(1/2) + 2sin^-1(1/2)

Sol:- cos^-1(1/2) + 2sin^-1(1/2) ………………..(1)
Let A = cos^-1(1/2) ………………..(2)

⇒ cos A = 1/2

⇒ cos A = cos (Π/3)

⇒ A = Π/3 { lies in range of cos^-1x [0,Π] }

Let B = sin^-1(1/2) ………………..(3)

⇒ sin B = 1/2

⇒ sin B = sin (Π/6)

⇒ B = Π/6 { lies in range of sin^-1x [-Π/2,Π/2] }

put Eq. 2,3 values in Eq. 1:-

= A+2B

put A, B values:-

= Π/3+2Π/6

= 2Π/3 (Ans……)

13. If sin^-1x = y , then

(A) 0≤ y ≤ Π

(B) -Π/2 ≤ y ≤ Π/2

(C) 0 < y < Π

(D) -Π/2 < y < Π/2

Sol:- We know value of sin^-1x lies in range of [-Π/2, Π/2]

then value of y lies between [-Π/2, Π/2].

(Ans…… B)

14. tan^-1(√3) - sec^-1(-2)

(A) Π

(B) - Π/3

(C) Π/3

(D) 2Π/3

Sol:- tan^-1(√3) - sec^-1(-2) ………………..(1)
Let A = tan^-1(√3) ………………..(2)

⇒ tan A = √3

⇒ tan A = tan (Π/3)

⇒ A = Π/3 { lies in range of tan^-1x (-Π/2,Π/2) }

Let B = sec^-1(-2) ………………..(3)

We know sec^-1(-x) = Π-sec^-1(x)

⇒ B = Π - sec^-1(2)

⇒ sec^-1(2) = Π - B

⇒ sec (Π-B) = 2

⇒ sec (Π-B) = sec(Π/3)

⇒ Π - B = Π/3

⇒ B = Π - Π/3

⇒ B = 2Π/3 { lies in range of sec^-1x [0,Π]-{ Π/2} }

put Eq. 2,3 values in Eq. 1:-

= A-B

put A, B values:-

= Π/3 – 2Π/3

= -Π/3

(Ans……B)

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