Question 1
Find the principal value of sin⁻¹(−1/2).
Steps only
sin(−π/6) = −1/2
sin⁻¹(−1/2) = −π/6
Answer
−π/6
CLASS 12 · MATHEMATICS PART I · CHAPTER 2
Inverse trigonometric functions
Read the short working first. Choose “Show explanation” on any answer for a clear explanation.
43 worked answersAll 2 exercise sets
43 worked answers · short steps first, explanations when you need them
Exercise 2.1
Textbook pages 26–27
Question 2
Find the principal value of cos⁻¹(√3/2).
Steps only
cos(π/6) = √3/2
cos⁻¹(√3/2) = π/6
Answer
π/6
Question 3
Find the principal value of cosec⁻¹(2).
Steps only
sin θ = 1/2
cosec⁻¹(2) = π/6
Answer
π/6
Question 4
Find the principal value of tan⁻¹(−√3).
Steps only
tan(−π/3) = −√3
tan⁻¹(−√3) = −π/3
Answer
−π/3
Question 5
Find the principal value of cos⁻¹(−1/2).
Steps only
cos(2π/3) = −1/2
cos⁻¹(−1/2) = 2π/3
Answer
2π/3
Question 6
Find the principal value of tan⁻¹(−1).
Steps only
tan(−π/4) = −1
tan⁻¹(−1) = −π/4
Answer
−π/4
Question 7
Find the principal value of sec⁻¹(2/√3).
Steps only
cos θ = √3/2
sec⁻¹(2/√3) = π/6
Answer
π/6
Question 8
Find the principal value of cot⁻¹(√3).
Steps only
cot(π/6) = √3
cot⁻¹(√3) = π/6
Answer
π/6
Question 9
Find the principal value of cos⁻¹(−1/√2).
Steps only
cos(3π/4) = −1/√2
cos⁻¹(−1/√2) = 3π/4
Answer
3π/4
Question 10
Find the principal value of cosec⁻¹(−√2).
Steps only
sin θ = −1/√2
cosec⁻¹(−√2) = −π/4
Answer
−π/4
Question 11
Evaluate tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
Steps only
= π/4 + 2π/3 − π/6
= (3 + 8 − 2)π/12
= 3π/4
Answer
3π/4
Question 12
Evaluate cos⁻¹(1/2) + 2sin⁻¹(1/2).
Steps only
= π/3 + 2(π/6)
= 2π/3
Answer
2π/3
Question 13
If sin⁻¹x = y, which interval must contain y?
Steps only
−π/2 ≤ y ≤ π/2
Answer
Option B: [−π/2, π/2].
Question 14
Evaluate tan⁻¹√3 − sec⁻¹(−2).
Steps only
tan⁻¹√3 = π/3
sec⁻¹(−2) = 2π/3
π/3 − 2π/3 = −π/3
Answer
Option B: −π/3.
Exercise 2.2
Textbook pages 29–30
Question 1
Prove 3sin⁻¹x = sin⁻¹(3x − 4x³) for −1/2 ≤ x ≤ 1/2.
Steps only
θ = sin⁻¹x; −π/6 ≤ θ ≤ π/6
3x − 4x³ = sin 3θ
sin⁻¹(sin 3θ) = 3θ = 3sin⁻¹x
Answer
The identity holds on the given interval.
Question 2
Prove 3cos⁻¹x = cos⁻¹(4x³ − 3x) for 1/2 ≤ x ≤ 1.
Steps only
θ = cos⁻¹x; 0 ≤ θ ≤ π/3
4x³ − 3x = cos 3θ
cos⁻¹(cos 3θ) = 3θ = 3cos⁻¹x
Answer
The identity holds on the given interval.
Question 3
Simplify tan⁻¹((√(1+x²) − 1)/x), x ≠ 0.
Steps only
θ = tan⁻¹x; sec θ = √(1+x²)
(√(1+x²) − 1)/x = (sec θ − 1)/tan θ = tan(θ/2)
= θ/2 = (1/2)tan⁻¹x
Answer
(1/2)tan⁻¹x.
Question 4
Simplify tan⁻¹√((1−cos x)/(1+cos x)), 0 < x < π.
Steps only
(1−cos x)/(1+cos x) = tan²(x/2)
√(tan²(x/2)) = tan(x/2)
tan⁻¹(tan(x/2)) = x/2
Answer
x/2.
Question 5
Simplify tan⁻¹((cos x−sin x)/(cos x+sin x)), −π/4 < x < 3π/4.
Steps only
(cos x−sin x)/(cos x+sin x) = tan(π/4−x)
−π/2 < π/4−x < π/2
= π/4−x
Answer
π/4 − x.
Question 6
Simplify tan⁻¹(x/√(a²−x²)), |x| < a.
Steps only
θ = sin⁻¹(x/a)
tan θ = x/√(a²−x²)
= sin⁻¹(x/a)
Answer
sin⁻¹(x/a).
Question 7
Simplify tan⁻¹((3a²x−x³)/(a³−3ax²)), a > 0 and −a/√3 < x < a/√3.
Steps only
θ = tan⁻¹(x/a); −π/6 < θ < π/6
tan 3θ = (3a²x−x³)/(a³−3ax²)
= 3θ = 3tan⁻¹(x/a)
Answer
3tan⁻¹(x/a).
Question 8
Evaluate tan⁻¹[2cos(2sin⁻¹(1/2))].
Steps only
sin⁻¹(1/2) = π/6
2cos(π/3) = 1
tan⁻¹1 = π/4
Answer
π/4
Question 9
Evaluate tan{(1/2)[sin⁻¹(2x/(1+x²)) + cos⁻¹((1−y²)/(1+y²))]}, |x| < 1, y > 0, xy < 1.
Steps only
sin⁻¹(2x/(1+x²)) = 2tan⁻¹x
cos⁻¹((1−y²)/(1+y²)) = 2tan⁻¹y
tan(tan⁻¹x + tan⁻¹y) = (x+y)/(1−xy)
Answer
(x+y)/(1−xy).
Question 10
Evaluate sin⁻¹(sin(2π/3)).
Steps only
sin(2π/3) = sin(π/3)
= π/3
Answer
π/3
Question 11
Evaluate tan⁻¹(tan(3π/4)).
Steps only
tan(3π/4) = −1
= −π/4
Answer
−π/4
Question 12
Evaluate tan(sin⁻¹(3/5) + cot⁻¹(3/2)).
Steps only
tan A = 3/4; tan B = 2/3
tan(A+B) = (3/4+2/3)/(1−(3/4)(2/3))
= 17/6
Answer
17/6
Question 13
Evaluate cos⁻¹(cos(7π/6)).
Steps only
cos(7π/6) = −√3/2
= 5π/6
Answer
Option B: 5π/6.
Question 14
Evaluate sin(π/3 − sin⁻¹(−1/2)).
Steps only
sin⁻¹(−1/2) = −π/6
sin(π/3 + π/6) = sin(π/2) = 1
Answer
Option D: 1.
Question 15
Evaluate tan⁻¹√3 − cot⁻¹(−√3).
Steps only
tan⁻¹√3 = π/3
cot⁻¹(−√3) = 5π/6
π/3 − 5π/6 = −π/2
Answer
Option B: −π/2.
Miscellaneous Exercise
Textbook pages 31
Question 1
Evaluate cos⁻¹(cos(13π/6)).
Steps only
13π/6 = 2π + π/6
cos⁻¹(cos(13π/6)) = π/6
Answer
π/6
Question 2
Evaluate tan⁻¹(tan(7π/6)).
Steps only
7π/6 = π + π/6
tan⁻¹(tan(7π/6)) = π/6
Answer
π/6
Question 3
Prove 2sin⁻¹(3/5) = tan⁻¹(24/7).
Steps only
A = sin⁻¹(3/5); tan A = 3/4
tan 2A = (2×3/4)/(1−9/16) = 24/7
2A = tan⁻¹(24/7)
Answer
Identity proved.
Question 4
Prove sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).
Steps only
sin A = 8/17; cos A = 15/17
sin B = 3/5; cos B = 4/5
sin(A+B) = 77/85; cos(A+B) = 36/85
tan(A+B) = 77/36
Answer
Identity proved.
Question 5
Prove cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65).
Steps only
cos A = 4/5; sin A = 3/5
cos B = 12/13; sin B = 5/13
cos(A+B) = (48−15)/65 = 33/65
Answer
Identity proved.
Question 6
Prove cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65).
Steps only
sin A = 5/13; cos A = 12/13
sin B = 3/5; cos B = 4/5
sin(A+B) = (20+36)/65 = 56/65
cos(A+B) = (48−15)/65 = 33/65 > 0
Answer
Identity proved.
Question 7
Prove tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5).
Steps only
tan A = 5/12; tan B = 4/3
tan(A+B) = (5/12+4/3)/(1−(5/12)(4/3)) = 63/16
cos(A+B) = (36−20)/65 = 16/65 > 0
Answer
Identity proved.
Question 8
Prove tan⁻¹√x = (1/2)cos⁻¹((1−x)/(1+x)), 0 ≤ x ≤ 1.
Steps only
θ = tan⁻¹√x; 0 ≤ θ ≤ π/4
cos 2θ = (1−tan²θ)/(1+tan²θ) = (1−x)/(1+x)
cos⁻¹((1−x)/(1+x)) = 2θ
Answer
Identity proved.
Question 9
Prove cot⁻¹((√(1+sin x)+√(1−sin x))/(√(1+sin x)−√(1−sin x))) = x/2 for 0 < x < π/4.
Steps only
√(1+sin x) = cos(x/2)+sin(x/2)
√(1−sin x) = cos(x/2)−sin(x/2)
ratio = 2cos(x/2)/(2sin(x/2)) = cot(x/2)
Answer
x/2.
Question 10
Prove tan⁻¹((√(1+x)−√(1−x))/(√(1+x)+√(1−x))) = π/4 − (1/2)cos⁻¹x, −1/√2 ≤ x ≤ 1.
Steps only
θ = (1/2)cos⁻¹x; 0 ≤ θ ≤ 3π/8
√(1+x) = √2 cos θ; √(1−x) = √2 sin θ
ratio = (cos θ−sin θ)/(cos θ+sin θ) = tan(π/4−θ)
Answer
Identity proved.
Question 11
Solve 2tan⁻¹(cos x) = tan⁻¹(2cosec x).
Steps only
sin x ≠ 0
2cos x/(1−cos²x) = 2/sin x
cos x = sin x
x = π/4 + kπ, k ∈ ℤ
Answer
x = π/4 + kπ, for any integer k.
Question 12
Solve tan⁻¹((1−x)/(1+x)) = (1/2)tan⁻¹x, x > 0.
Steps only
θ = tan⁻¹x; 0 < θ < π/2
tan⁻¹((1−x)/(1+x)) = π/4−θ
π/4−θ = θ/2
θ = π/6; x = 1/√3
Answer
x = 1/√3.
Question 13
Evaluate sin(tan⁻¹x), |x| < 1.
Steps only
θ = tan⁻¹x; cos θ = 1/√(1+x²)
sin θ = tan θ cos θ = x/√(1+x²)
Answer
Option D: x/√(1+x²).
Question 14
Solve sin⁻¹(1−x) − 2sin⁻¹x = π/2.
Steps only
0 ≤ x ≤ 1
sin⁻¹(1−x) ≤ π/2; 2sin⁻¹x ≥ 0
Equality requires sin⁻¹x = 0
x = 0
Answer
Option C: x = 0.