Exercise 2.1

Question 1: Find the principal value of cosec⁻¹ (− 1/ 2)

Answer 1:

Let cosec⁻¹ (− 1/2) = y, Then

cosec y = − 1/2 = −cosec (π/6) = cosec (− π/6)

We know that the range of the principal value branch of sin⁻¹ is [-(π/2) , (π/2) ] and

sin (-(π/6)) = - 1/2

Therefore, the principal value of sin^-1 (-(1/2)) is – (π/6) .

 

Question 2: Find the principal value of cos −1

Answer 2:

Let , then cos y =  = cosec  (π/6)

We know that the range of the principal value branch of cos−1 is [0, π] and cos (π/6) = Therefore, the principal value of cos^-1  is (π/6).

 

Question 3: Find the principal value of cosec⁻¹ (2).

Answer 3:

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).

Question 4: Find the principal value of tan⁻¹ (−).

Answer 4:

Let tan⁻¹ (−) = y, then tan y = = -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)) =

 Therefore, the principal value of tan^-1 () is – (π/3).

Question 5: Find the principal value of cos −1 (− 1 2 ).

Answer 5:

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/3).

 

Question 6: Find the principal value of tan⁻¹ (−1).

Answer 6: 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).

Question 7: Find the principal value of sec⁻¹ (2/ ).

Answer 7:

Let sec⁻¹ (2/) = y, then sec y = (2/) = 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/).

 Therefore, the principal value of sec⁻¹ (2/) is (π/6).

Question 8: Find the principal value of cot⁻¹.

Answer 8:

Let cot^-1 = y, then cot y = = cot (π/6).

We know that the range of the principal value branch of cot⁻¹ is (0, π) and cot (π/6) = .

Therefore, the principal value of cot^-1 is (π/6).

Question 9: Find the principal value of cos ⁻¹ (−(1/)).

Answer 9:

Let cos^-1 (- (1/)) = y, then

cos y = -(1/) = -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 (3π/4) = -(1/).

 Therefore, the principal value of cos^-1 (- 1/) is (3π/4).

Question 10: Find the principal value of cosec⁻¹ (−).

Answer 10:

Let cosec⁻¹ (−) = y, then

cosec y = − = −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)) = -.

Therefore, the principal value of cosec^-1 (-) is –(π/4).

Question 11: Find the value of tan⁻¹ (1) + cos ⁻¹ (−(1/2)) + sin⁻¹ (−(1/2)).

Answer 11:

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)= −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].

∴ 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 [-(π/2),(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) = (π/4)

Question 12: Find the value of cos ⁻¹ (1/2) + 2sin⁻¹ (1/2)

Answer 12:

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).

Question 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 13:

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.

Question 14: tan−1√3 − sec−1 (−2) is equal to

(A) π

(B) –(π/3)

(C) (π/3)

(D) (2π/3)

Answer 14:

Let tan⁻¹ = x, then

tan x = = tan (π/3)

We know that the range of the principal value branch of tan⁻¹ is (-(π/2),(π/2)).

∴ tan^-1 = (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 -sec^-1 (-2) = (π/3) – (2π/3) = - (π/3)

Hence, the option (B) is correct.