Class 12 Mathematics - NCERT Mathematics Part I
NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions
Chapter 2 introduces the inverse of trigonometric functions by restricting their domains. Students learn to find principal values of sin⁻¹, cos⁻¹, tan⁻¹, cosec⁻¹, sec⁻¹, and cot⁻¹, and to apply key identities involving these functions. This chapter is part of the Relations and Functions unit which carries 8 marks in the CBSE Class 12 board exam 2027. Inverse trigonometric functions typically contribute about 4 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 2 has 2 exercises plus a Miscellaneous Exercise. Exercise 2.1 (14 questions) covers principal values. Exercise 2.2 (21 questions) covers properties, proofs, and simplifications. The Miscellaneous Exercise (17 questions) combines all concepts with identity proofs.
Table of Contents
2.1Exercise 2.1 - Principal Values of Inverse Trigonometric Functions
Q1Find the principal value of sin⁻¹(−1/2).
Let sin⁻¹(−1/2) = y. Then sin y = −1/2.
We need to find y such that sin y = −1/2 and y ∈ [−π/2, π/2] (the principal value branch of sin⁻¹).
We know that sin(π/6) = 1/2.
Using the property sin(−θ) = −sin θ:
sin(−π/6) = −sin(π/6) = −1/2 ✓
Check: −π/6 ∈ [−π/2, π/2]? Yes, since −π/2 ≤ −π/6 ≤ π/2. ✓
∴ Principal value of sin⁻¹(−1/2) = −π/6
Q2Find the principal value of cos⁻¹(√3/2).
Let cos⁻¹(√3/2) = y. Then cos y = √3/2.
We need to find y such that cos y = √3/2 and y ∈ [0, π] (the principal value branch of cos⁻¹).
From the standard angle table: cos(π/6) = √3/2.
Check: π/6 ∈ [0, π]? Yes. ✓
∴ Principal value of cos⁻¹(√3/2) = π/6
Q3Find the principal value of cosec⁻¹(2).
Let cosec⁻¹(2) = y. Then cosec y = 2.
Since cosec y = 1/sin y, we have sin y = 1/cosec y = 1/2.
We need y such that sin y = 1/2 and y ∈ [−π/2, π/2] − {0} (the principal value branch of cosec⁻¹).
From the standard angle table: sin(π/6) = 1/2.
Check: π/6 ∈ [−π/2, π/2] − {0}? Yes (π/6 ≠ 0 and π/6 is within [−π/2, π/2]). ✓
∴ Principal value of cosec⁻¹(2) = π/6
Q4Find the principal value of tan⁻¹(−√3).
Let tan⁻¹(−√3) = y. Then tan y = −√3.
We need y such that tan y = −√3 and y ∈ (−π/2, π/2) (the principal value branch of tan⁻¹).
From the standard angle table: tan(π/3) = √3.
Using the property tan(−θ) = −tan θ:
tan(−π/3) = −tan(π/3) = −√3 ✓
Check: −π/3 ∈ (−π/2, π/2)? Yes, since −π/2 < −π/3 < π/2. ✓
∴ Principal value of tan⁻¹(−√3) = −π/3
Q5Find the principal value of cos⁻¹(−1/2).
Let cos⁻¹(−1/2) = y. Then cos y = −1/2.
We need y such that cos y = −1/2 and y ∈ [0, π] (the principal value branch of cos⁻¹).
Since the value is negative, the angle must be in the second quadrant (between π/2 and π).
Using the property cos(π − θ) = −cos θ:
cos(π − π/3) = −cos(π/3) = −1/2
So cos(2π/3) = −1/2. ✓
Check: 2π/3 ∈ [0, π]? Yes. ✓
∴ Principal value of cos⁻¹(−1/2) = 2π/3
Key insight: For cos⁻¹ of negative values, use the identity cos⁻¹(−x) = π − cos⁻¹(x). Here: cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3.
Q6Find the principal value of tan⁻¹(−1).
Let tan⁻¹(−1) = y. Then tan y = −1.
We need y such that tan y = −1 and y ∈ (−π/2, π/2) (the principal value branch of tan⁻¹).
From the standard angle table: tan(π/4) = 1.
Using the property tan(−θ) = −tan θ:
tan(−π/4) = −tan(π/4) = −1 ✓
Check: −π/4 ∈ (−π/2, π/2)? Yes. ✓
∴ Principal value of tan⁻¹(−1) = −π/4
Key insight: For tan⁻¹ of negative values, use tan⁻¹(−x) = −tan⁻¹(x). Here: tan⁻¹(−1) = −tan⁻¹(1) = −π/4.
Q7Find the principal value of sec⁻¹(2/√3).
Let sec⁻¹(2/√3) = y. Then sec y = 2/√3.
Since sec y = 1/cos y, we have cos y = √3/2.
We need y such that cos y = √3/2 and y ∈ [0, π] − {π/2} (the principal value branch of sec⁻¹).
From the standard angle table: cos(π/6) = √3/2.
Check: π/6 ∈ [0, π] − {π/2}? Yes (π/6 ≠ π/2 and 0 ≤ π/6 ≤ π). ✓
∴ Principal value of sec⁻¹(2/√3) = π/6
Q8Find the principal value of cot⁻¹(√3).
Let cot⁻¹(√3) = y. Then cot y = √3.
Since cot y = 1/tan y, we have tan y = 1/√3.
We need y such that tan y = 1/√3 and y ∈ (0, π) (the principal value branch of cot⁻¹).
From the standard angle table: tan(π/6) = 1/√3.
Check: π/6 ∈ (0, π)? Yes. ✓
∴ Principal value of cot⁻¹(√3) = π/6
Note: The range of cot⁻¹ is (0, π), which is different from tan⁻¹ whose range is (−π/2, π/2). This is a common source of confusion.
Q9Find the principal value of cos⁻¹(−1/√2).
Let cos⁻¹(−1/√2) = y. Then cos y = −1/√2.
We need y such that cos y = −1/√2 and y ∈ [0, π] (the principal value branch of cos⁻¹).
Since the value is negative, the angle is in the second quadrant.
Using cos(π − θ) = −cos θ:
cos(π − π/4) = −cos(π/4) = −1/√2
So cos(3π/4) = −1/√2. ✓
Check: 3π/4 ∈ [0, π]? Yes. ✓
∴ Principal value of cos⁻¹(−1/√2) = 3π/4
Alternatively: cos⁻¹(−1/√2) = π − cos⁻¹(1/√2) = π − π/4 = 3π/4.
Q10Find the principal value of cosec⁻¹(−√2).
Let cosec⁻¹(−√2) = y. Then cosec y = −√2.
Since cosec y = 1/sin y, we have sin y = −1/√2.
We need y such that sin y = −1/√2 and y ∈ [−π/2, π/2] − {0} (the principal value branch of cosec⁻¹).
Since the value is negative, y must be in [−π/2, 0).
From the standard angle table: sin(π/4) = 1/√2.
Using sin(−θ) = −sin θ:
sin(−π/4) = −1/√2. ✓
Check: −π/4 ∈ [−π/2, π/2] − {0}? Yes (−π/4 ≠ 0 and −π/2 ≤ −π/4 ≤ π/2). ✓
∴ Principal value of cosec⁻¹(−√2) = −π/4
Q11Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
tan⁻¹(1) = π/4 (since tan(π/4) = 1)
cos⁻¹(−1/2) = 2π/3 (since cos(2π/3) = −1/2, and 2π/3 ∈ [0, π])
sin⁻¹(−1/2) = −π/6 (since sin(−π/6) = −1/2, and −π/6 ∈ [−π/2, π/2])
Sum = π/4 + 2π/3 + (−π/6) = π/4 + 2π/3 − π/6
LCM of 4, 3, 6 = 12:
= 3π/12 + 8π/12 − 2π/12
= 9π/12
= 3π/4
Q12Find the value of cos⁻¹(1/2) + 2sin⁻¹(1/2).
cos⁻¹(1/2) = π/3 (since cos(π/3) = 1/2)
sin⁻¹(1/2) = π/6 (since sin(π/6) = 1/2)
Sum = π/3 + 2(π/6) = π/3 + π/3 = 2π/3
Q13If 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
The principal value branch of sin⁻¹ is [−π/2, π/2].
This means −π/2 ≤ y ≤ π/2 (the endpoints are included).
Answer: (B) −π/2 ≤ y ≤ π/2
Q14tan⁻¹(√3) − sec⁻¹(−2) is equal to (A) π (B) −π/3 (C) π/3 (D) 2π/3
tan⁻¹(√3) = π/3 (since tan(π/3) = √3)
sec⁻¹(−2): sec y = −2 means cos y = −1/2. cos(2π/3) = −1/2, and 2π/3 ∈ [0, π] − {π/2}.
So sec⁻¹(−2) = 2π/3.
tan⁻¹(√3) − sec⁻¹(−2) = π/3 − 2π/3 = −π/3
Answer: (B) −π/3
Stuck on a question? AI explains any step in detail - in English or Hindi.
Ask AI TutorPrevious Year Board Exam Questions
Find the value of sin⁻¹(sin(2π/3)).
2π/3 is NOT in [−π/2, π/2] (the principal branch of sin⁻¹).
sin(2π/3) = sin(π − π/3) = sin(π/3) = √3/2.
So sin⁻¹(sin(2π/3)) = sin⁻¹(√3/2) = π/3
(Note: sin⁻¹(sin x) = x only when x ∈ [−π/2, π/2]. Otherwise, find the equivalent angle in the principal branch.)
Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
tan⁻¹(1) = π/4
cos⁻¹(−1/2) = 2π/3
sin⁻¹(−1/2) = −π/6
Sum = π/4 + 2π/3 − π/6 = 3π/12 + 8π/12 − 2π/12 = 3π/4
Prove that 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17).
First: 2tan⁻¹(1/2) = tan⁻¹(2 × (1/2)/(1 − (1/2)²)) = tan⁻¹(1/(1 − 1/4)) = tan⁻¹(1/(3/4)) = tan⁻¹(4/3)
Now: tan⁻¹(4/3) + tan⁻¹(1/7) = tan⁻¹((4/3 + 1/7)/(1 − 4/3 × 1/7))
= tan⁻¹((28/21 + 3/21)/(1 − 4/21))
= tan⁻¹((31/21)/(17/21))
= tan⁻¹(31/17)
Hence 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17). ∎
Write the domain of the function f(x) = sin⁻¹(x²) + cos⁻¹(x²).
sin⁻¹(x²) + cos⁻¹(x²) = π/2 (using the identity sin⁻¹(t) + cos⁻¹(t) = π/2)
But for this to be defined, we need x² ∈ [−1, 1], i.e., 0 ≤ x² ≤ 1.
So −1 ≤ x ≤ 1.
Domain = [−1, 1]
Simplify: tan⁻¹(cos x/(1 − sin x)), −3π/2 < x < π/2.
We need to simplify tan⁻¹(cos x/(1 − sin x)) for −3π/2 < x < π/2.
Step 1: Express cos x and sin x using half-angle identities:
cos x = cos²(x/2) − sin²(x/2)
1 − sin x = 1 − 2sin(x/2)cos(x/2) = cos²(x/2) + sin²(x/2) − 2sin(x/2)cos(x/2) = (cos(x/2) − sin(x/2))²
Step 2: Substitute:
cos x/(1 − sin x) = (cos²(x/2) − sin²(x/2))/(cos(x/2) − sin(x/2))²
= (cos(x/2) + sin(x/2))(cos(x/2) − sin(x/2))/(cos(x/2) − sin(x/2))²
= (cos(x/2) + sin(x/2))/(cos(x/2) − sin(x/2))
Step 3: Divide numerator and denominator by cos(x/2):
= (1 + tan(x/2))/(1 − tan(x/2))
= (tan(π/4) + tan(x/2))/(1 − tan(π/4)·tan(x/2)) [since tan(π/4) = 1]
= tan(π/4 + x/2) [using tan addition formula]
Step 4: Apply tan⁻¹:
tan⁻¹(cos x/(1 − sin x)) = tan⁻¹(tan(π/4 + x/2)) = π/4 + x/2
(This is valid since for −3π/2 < x < π/2, we have −π/2 < π/4 + x/2 < π/2, which is within the principal branch of tan⁻¹.)
Find the value of sin⁻¹(sin(3π/5)).
3π/5 ∉ [−π/2, π/2] (principal branch of sin⁻¹).
sin(3π/5) = sin(π − 3π/5) = sin(2π/5).
Since 2π/5 ∈ [−π/2, π/2] (2π/5 ≈ 1.257 < π/2 ≈ 1.571),
sin⁻¹(sin(3π/5)) = 2π/5
Prove that tan⁻¹(1/5) + tan⁻¹(1/7) + tan⁻¹(1/3) + tan⁻¹(1/8) = π/4.
Group the terms:
tan⁻¹(1/5) + tan⁻¹(1/7) = tan⁻¹((1/5 + 1/7)/(1 − 1/35)) = tan⁻¹((12/35)/(34/35)) = tan⁻¹(12/34) = tan⁻¹(6/17)
tan⁻¹(1/3) + tan⁻¹(1/8) = tan⁻¹((1/3 + 1/8)/(1 − 1/24)) = tan⁻¹((11/24)/(23/24)) = tan⁻¹(11/23)
Now: tan⁻¹(6/17) + tan⁻¹(11/23) = tan⁻¹((6/17 + 11/23)/(1 − 6/17 × 11/23))
= tan⁻¹((138/391 + 187/391)/(1 − 66/391))
= tan⁻¹((325/391)/(325/391))
= tan⁻¹(1) = π/4 ∎
Write the value of cos⁻¹(cos(7π/6)).
7π/6 ∉ [0, π] (principal branch of cos⁻¹).
cos(7π/6) = cos(2π − 5π/6) = cos(5π/6).
Since 5π/6 ∈ [0, π],
cos⁻¹(cos(7π/6)) = 5π/6
Prove that 3cos⁻¹(x) = cos⁻¹(4x³ − 3x), x ∈ [1/2, 1].
Let cos⁻¹(x) = θ, so x = cos θ where θ ∈ [0, π/3] (since x ∈ [1/2, 1]).
RHS = cos⁻¹(4cos³θ − 3cos θ) = cos⁻¹(cos 3θ)
Since θ ∈ [0, π/3], we have 3θ ∈ [0, π] (within principal branch of cos⁻¹).
So cos⁻¹(cos 3θ) = 3θ = 3cos⁻¹(x) = LHS. ∎
Simplify: tan⁻¹(√(1 + x²) − 1)/x), x ≠ 0.
Step 1: Substitute x = tan θ (where θ = tan⁻¹(x)).
Then √(1 + x²) = √(1 + tan²θ) = √(sec²θ) = sec θ (taking positive root since sec θ > 0 for θ ∈ (−π/2, π/2)).
Step 2: Simplify the expression inside tan⁻¹:
(√(1 + x²) − 1)/x = (sec θ − 1)/tan θ
= (1/cos θ − 1)/(sin θ/cos θ)
= (1 − cos θ)/(sin θ)
Step 3: Use half-angle identities:
1 − cos θ = 2sin²(θ/2)
sin θ = 2sin(θ/2)cos(θ/2)
So (1 − cos θ)/sin θ = 2sin²(θ/2)/(2sin(θ/2)cos(θ/2)) = sin(θ/2)/cos(θ/2) = tan(θ/2)
Step 4: Apply tan⁻¹:
tan⁻¹((√(1 + x²) − 1)/x) = tan⁻¹(tan(θ/2)) = θ/2
Since θ = tan⁻¹(x):
∴ tan⁻¹((√(1 + x²) − 1)/x) = (1/2)tan⁻¹(x)
Prove that tan⁻¹(1/2) + tan⁻¹(1/5) + tan⁻¹(1/8) = π/4.
tan⁻¹(1/2) + tan⁻¹(1/5) = tan⁻¹((1/2 + 1/5)/(1 − 1/10)) = tan⁻¹((7/10)/(9/10)) = tan⁻¹(7/9)
Now: tan⁻¹(7/9) + tan⁻¹(1/8) = tan⁻¹((7/9 + 1/8)/(1 − 7/72))
= tan⁻¹((56/72 + 9/72)/(72/72 − 7/72))
= tan⁻¹((65/72)/(65/72))
= tan⁻¹(1) = π/4 ∎
Find the principal value of tan⁻¹(tan(3π/4)).
3π/4 ∉ (−π/2, π/2) (principal branch of tan⁻¹).
tan(3π/4) = tan(π − π/4) = −tan(π/4) = −1.
tan⁻¹(−1) = −π/4 (since −π/4 ∈ (−π/2, π/2)).
tan⁻¹(tan(3π/4)) = −π/4
Prove that sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).
Step 1: Let α = sin⁻¹(8/17) and β = sin⁻¹(3/5).
Then sin α = 8/17 and sin β = 3/5.
Step 2: Find cos α and cos β using sin²θ + cos²θ = 1:
cos α = √(1 − sin²α) = √(1 − 64/289) = √(225/289) = 15/17
cos β = √(1 − sin²β) = √(1 − 9/25) = √(16/25) = 4/5
(Taking positive roots since α, β ∈ [0, π/2] as sin values are positive.)
Step 3: Find sin(α + β) using the addition formula:
sin(α + β) = sin α cos β + cos α sin β
= (8/17)(4/5) + (15/17)(3/5)
= 32/85 + 45/85
= 77/85
Step 4: Find cos(α + β):
cos(α + β) = cos α cos β − sin α sin β
= (15/17)(4/5) − (8/17)(3/5)
= 60/85 − 24/85
= 36/85
Step 5: Find tan(α + β):
tan(α + β) = sin(α + β)/cos(α + β) = (77/85)/(36/85) = 77/36
Therefore α + β = tan⁻¹(77/36).
∴ sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36) ∎
Find the value of tan⁻¹[2cos(2sin⁻¹(1/2))].
sin⁻¹(1/2) = π/6
2sin⁻¹(1/2) = 2 × π/6 = π/3
cos(π/3) = 1/2
2cos(π/3) = 2 × 1/2 = 1
tan⁻¹(1) = π/4
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The principal value of sin⁻¹(sin(5π/6)) is 5π/6.
Reason (R): sin⁻¹(sin x) = x for all x ∈ R.
Answer: (D)
Assertion is FALSE: 5π/6 ∉ [−π/2, π/2], so sin⁻¹(sin(5π/6)) = sin⁻¹(1/2) = π/6, not 5π/6. Reason is FALSE: sin⁻¹(sin x) = x only when x ∈ [−π/2, π/2].
AR2
Assertion (A): sin⁻¹(1/2) + cos⁻¹(1/2) = π/2.
Reason (R): sin⁻¹(x) + cos⁻¹(x) = π/2 for all x ∈ [−1, 1].
Answer: (A)
Both true. sin⁻¹(1/2) = π/6, cos⁻¹(1/2) = π/3, sum = π/2. The reason is the general identity and correctly explains the assertion.
AR3
Assertion (A): The range of cos⁻¹(x) is [0, π].
Reason (R): Cosine function is restricted to [0, π] to make it one-one and onto [−1, 1].
Answer: (A)
Both true. cos: [0, π] → [−1, 1] is bijective, so its inverse cos⁻¹: [−1, 1] → [0, π] exists. The reason correctly explains why the range is [0, π].
AR4
Assertion (A): tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = π.
Reason (R): tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1−xy)) for all x, y.
Answer: (C)
Assertion is TRUE: tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = π/4 + tan⁻¹(2) + tan⁻¹(3). Since 2×3 > 1, we use tan⁻¹(2) + tan⁻¹(3) = π + tan⁻¹((2+3)/(1−6)) = π + tan⁻¹(−1) = π − π/4 = 3π/4. Total = π/4 + 3π/4 = π. Reason is FALSE: the formula tan⁻¹(x)+tan⁻¹(y)=tan⁻¹((x+y)/(1−xy)) is valid only when xy < 1.
AR5
Assertion (A): sin⁻¹(−x) = −sin⁻¹(x) for all x ∈ [−1, 1].
Reason (R): sin⁻¹ is an odd function.
Answer: (A)
Both true. sin⁻¹(−x) = −sin⁻¹(x) follows from the fact that sin is an odd function. The reason correctly identifies sin⁻¹ as odd.
AR6
Assertion (A): cos⁻¹(−x) = π − cos⁻¹(x) for all x ∈ [−1, 1].
Reason (R): cos⁻¹ is an odd function.
Answer: (C)
Assertion is TRUE: cos⁻¹(−x) = π − cos⁻¹(x) is a standard identity. Reason is FALSE: cos⁻¹ is NOT an odd function. An odd function satisfies f(−x) = −f(x), but cos⁻¹(−x) = π − cos⁻¹(x) ≠ −cos⁻¹(x).
Case Study Questions
Two pilgrims are standing at different distances from a temple tower. The tower is 50 metres tall. Pilgrim A is standing at a distance such that the angle of elevation to the top of the tower is tan⁻¹(5/4). Pilgrim B is standing further away at a distance such that the angle of elevation is tan⁻¹(5/12).
(i) Find the distance of Pilgrim A from the base of the tower. (1 mark)
Angle of elevation = tan⁻¹(5/4), so tan θ = 5/4.
tan θ = height/distance = 50/d.
5/4 = 50/d → d = 50 × 4/5 = 40 metres
(ii) Find the distance of Pilgrim B from the base of the tower. (1 mark)
Angle of elevation = tan⁻¹(5/12), so tan θ = 5/12.
5/12 = 50/d → d = 50 × 12/5 = 120 metres
(iii)(a) Find the distance between Pilgrim A and Pilgrim B. (1 mark)
Distance = 120 − 40 = 80 metres (assuming both are on the same side).
(iii)(b) Find the value of tan⁻¹(5/4) − tan⁻¹(5/12). (1 mark)
tan⁻¹(5/4) − tan⁻¹(5/12) = tan⁻¹((5/4 − 5/12)/(1 + 5/4 × 5/12))
= tan⁻¹((15/12 − 5/12)/(1 + 25/48))
= tan⁻¹((10/12)/(73/48))
= tan⁻¹((10/12) × (48/73))
= tan⁻¹(40/73)
= tan⁻¹(40/73)
A rectangular billboard of height 4 metres is mounted on top of a wall that is 3 metres high. A person standing at a certain distance from the wall observes the bottom of the billboard at an angle of elevation of tan⁻¹(3/10) from the ground.
(i) What is the distance of the person from the wall? (1 mark)
The bottom of the billboard is at 3 metres height.
tan θ = 3/d where θ = tan⁻¹(3/10), so tan θ = 3/10.
3/10 = 3/d → d = 10 metres
(ii) What is the angle of elevation to the top of the billboard from where the person stands? (1 mark)
Top of billboard = 3 + 4 = 7 metres.
Angle of elevation = tan⁻¹(7/10) = tan⁻¹(7/10)
(iii) What is the angle subtended by the billboard at the person's eye (assuming eye level at ground)? (2 marks)
Angle subtended = tan⁻¹(7/10) − tan⁻¹(3/10)
= tan⁻¹((7/10 − 3/10)/(1 + 7/10 × 3/10))
= tan⁻¹((4/10)/(1 + 21/100))
= tan⁻¹((4/10)/(121/100))
= tan⁻¹(40/121)
= tan⁻¹(40/121)
MCQ Practice
MCQ1. The principal value of sin⁻¹(−√3/2) is:
(A) −2π/3
(B) −π/3
(C) π/3
(D) 2π/3
Answer: (B) - sin⁻¹(−√3/2) = −sin⁻¹(√3/2) = −π/3. Check: −π/3 ∈ [−π/2, π/2] ✓.
MCQ2. The value of cos⁻¹(cos(−π/3)) is:
(A) −π/3
(B) π/3
(C) 2π/3
(D) −2π/3
Answer: (B) - cos(−π/3) = cos(π/3) = 1/2 (cosine is even). cos⁻¹(1/2) = π/3 ∈ [0, π].
MCQ3. If tan⁻¹(x) = π/4 − tan⁻¹(1/3), then x is:
(A) 1/3
(B) 1/2
(C) 1/4
(D) 1/6
Answer: (B) - tan⁻¹(x) = tan⁻¹(1) − tan⁻¹(1/3) = tan⁻¹((1 − 1/3)/(1 + 1/3)) = tan⁻¹((2/3)/(4/3)) = tan⁻¹(1/2). So x = 1/2.
MCQ4. The domain of sin⁻¹(2x) is:
(A) [0, 1]
(B) [−1, 1]
(C) [−1/2, 1/2]
(D) [−2, 2]
Answer: (C) - For sin⁻¹(2x) to be defined, −1 ≤ 2x ≤ 1, so −1/2 ≤ x ≤ 1/2.
MCQ5. tan⁻¹(√3) − cot⁻¹(−√3) is equal to:
(A) π
(B) −π/2
(C) 0
(D) 2√3
Answer: (B) - tan⁻¹(√3) = π/3. cot⁻¹(−√3) = π − cot⁻¹(√3) = π − π/6 = 5π/6. So π/3 − 5π/6 = 2π/6 − 5π/6 = −3π/6 = −π/2.
MCQ6. sin(tan⁻¹(x)), |x| < 1, is equal to:
(A) x/√(1−x²)
(B) 1/√(1−x²)
(C) 1/√(1+x²)
(D) x/√(1+x²)
Answer: (D) - Let θ = tan⁻¹(x). Then tan θ = x. In a right triangle: opposite = x, adjacent = 1, hypotenuse = √(1+x²). So sin θ = x/√(1+x²).
MCQ7. The principal value of cos⁻¹(cos(4π/3)) is:
(A) 4π/3
(B) 2π/3
(C) −π/3
(D) π/3
Answer: (B) - 4π/3 ∉ [0, π] (principal branch of cos⁻¹). cos(4π/3) = cos(π + π/3) = −cos(π/3) = −1/2. So cos⁻¹(cos(4π/3)) = cos⁻¹(−1/2) = 2π/3.
MCQ8. If sin⁻¹(x) + sin⁻¹(y) = π/2, then cos⁻¹(x) + cos⁻¹(y) equals:
(A) π/2
(B) π
(C) 0
(D) 2π/3
Answer: (A) - cos⁻¹(x) = π/2 − sin⁻¹(x) and cos⁻¹(y) = π/2 − sin⁻¹(y). So cos⁻¹(x) + cos⁻¹(y) = π − (sin⁻¹(x) + sin⁻¹(y)) = π − π/2 = π/2.
MCQ9. The value of 2tan⁻¹(1/3) is:
(A) tan⁻¹(3/4)
(B) tan⁻¹(4/3)
(C) tan⁻¹(2/3)
(D) tan⁻¹(1/3)
Answer: (A) - 2tan⁻¹(1/3) = tan⁻¹(2 × 1/3 / (1 − 1/9)) = tan⁻¹((2/3)/(8/9)) = tan⁻¹((2/3) × (9/8)) = tan⁻¹(6/8) = tan⁻¹(3/4).
MCQ10. The number of real solutions of the equation tan⁻¹(x) + tan⁻¹(1/x) = π/2 is:
(A) 0
(B) 1
(C) 2
(D) infinitely many
Answer: (D) - For x > 0: tan⁻¹(x) + tan⁻¹(1/x) = π/2 (standard identity). This holds for ALL positive x. So infinitely many solutions (all x > 0). Note: for x < 0, the sum is −π/2, not π/2.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the principal value of cos⁻¹(−1/2).
cos⁻¹(−1/2) = 2π/3
Find the principal value of tan⁻¹(−1).
tan⁻¹(−1) = −π/4
Find the value of sin⁻¹(sin(2π/3)).
2π/3 ∉ [−π/2, π/2]. sin(2π/3) = √3/2. sin⁻¹(√3/2) = π/3.
Write the domain of cos⁻¹(2x − 1).
−1 ≤ 2x − 1 ≤ 1 → 0 ≤ x ≤ 1. Domain = [0, 1].
Find the value of tan⁻¹(1) + tan⁻¹(0) + tan⁻¹(−1).
π/4 + 0 + (−π/4) = 0
2-3 Mark Questions
Prove that tan⁻¹(1/2) + tan⁻¹(1/5) + tan⁻¹(1/8) = π/4.
tan⁻¹(1/2) + tan⁻¹(1/5) = tan⁻¹(7/9). Then tan⁻¹(7/9) + tan⁻¹(1/8) = tan⁻¹((7/9+1/8)/(1−7/72)) = tan⁻¹(65/65) = tan⁻¹(1) = π/4.
Simplify: tan⁻¹(cos x − sin x)/(cos x + sin x), 0 < x < π.
Divide numerator and denominator by cos x: tan⁻¹((1 − tan x)/(1 + tan x)) = tan⁻¹(tan(π/4 − x)) = π/4 − x.
If tan⁻¹(x − 1)/(x − 2) + tan⁻¹(x + 1)/(x + 2) = π/4, find x.
Using addition formula: ((x−1)(x+2) + (x+1)(x−2))/((x−2)(x+2) − (x−1)(x+1)) = 1. Simplify: (2x²−2)/(−2x²+5−1) = (2x²−2)/(−2x²+4). Wait: numerator = x²+x−2+x²−x−2 = 2x²−4. Denominator = (x²−4)−(x²−1) = −3. So tan of sum = (2x²−4)/(−3). Set = 1: 2x²−4 = −3, 2x² = 1, x² = 1/2, x = ±1/√2.
Prove that sin⁻¹(3/5) − sin⁻¹(8/17) = cos⁻¹(84/85).
Let α = sin⁻¹(3/5), β = sin⁻¹(8/17). Then sin α = 3/5, cos α = 4/5, sin β = 8/17, cos β = 15/17. sin(α−β) = sin α cos β − cos α sin β = 45/85 − 32/85 = 13/85. cos(α−β) = cos α cos β + sin α sin β = 60/85 + 24/85 = 84/85. So α − β = cos⁻¹(84/85). ∎
3-4 Mark Questions
Prove that 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17).
2tan⁻¹(1/2) = tan⁻¹(2×(1/2)/(1−1/4)) = tan⁻¹(4/3). Then tan⁻¹(4/3) + tan⁻¹(1/7) = tan⁻¹((4/3+1/7)/(1−4/21)) = tan⁻¹((31/21)/(17/21)) = tan⁻¹(31/17). ∎
Prove that 3cos⁻¹(x) = cos⁻¹(4x³ − 3x), x ∈ [1/2, 1].
Let θ = cos⁻¹(x), x = cos θ, θ ∈ [0, π/3]. RHS = cos⁻¹(4cos³θ − 3cos θ) = cos⁻¹(cos 3θ). Since 3θ ∈ [0, π], this equals 3θ = 3cos⁻¹(x) = LHS. ∎
Prove that sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).
Let α = sin⁻¹(8/17), β = sin⁻¹(3/5). sin(α+β) = (8/17)(4/5) + (15/17)(3/5) = 32/85 + 45/85 = 77/85. cos(α+β) = (15/17)(4/5) − (8/17)(3/5) = 60/85 − 24/85 = 36/85. tan(α+β) = 77/36. So sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36). ∎
Key Concepts
Important Formulas
Study Plan - Inverse Trigonometric Functions
Principal Values
Memorise all 6 domain-range tables. Solve Exercise 2.1 Q1-Q10
Principal Values (continued)
Solve Exercise 2.1 Q11-Q14. Practice sin⁻¹(sin x) type problems
Properties and Proofs
Learn all identities. Solve Exercise 2.2 Q1-Q10
Properties (continued)
Solve Exercise 2.2 Q11-Q21 + Miscellaneous Exercise
Mixed Practice
Solve 5 PYQs + 10 MCQs. Focus on identity-based proofs
Common Mistakes to Avoid
- ✗Forgetting principal value ranges - sin⁻¹(sin(2π/3)) ≠ 2π/3 because 2π/3 is outside [−π/2, π/2]
- ✗Confusing sin⁻¹(x) with 1/sin(x) - sin⁻¹ means inverse function, not reciprocal
- ✗Using tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1−xy)) without checking the condition xy < 1
- ✗Not converting negative arguments properly - cos⁻¹(−x) = π − cos⁻¹(x), not −cos⁻¹(x)
- ✗Leaving answers in degrees instead of radians - all inverse trig values should be in radians
- ✗Forgetting that cot⁻¹(x) range is (0, π), not (−π/2, π/2) like tan⁻¹
Tips for Board Exam
- ✓Memorise all six principal value ranges - this is tested in MCQs every year
- ✓For sin⁻¹(sin x) type questions, first check if x is in the principal branch. If not, find the equivalent angle.
- ✓The three complementary identities (sin⁻¹+cos⁻¹, tan⁻¹+cot⁻¹, sec⁻¹+cosec⁻¹ = π/2) solve many MCQs instantly
- ✓For proof questions, use the tan⁻¹ addition formula: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1−ab))
- ✓Draw a right triangle to convert between inverse trig functions - if sin⁻¹(x) = θ, then cos θ = √(1−x²)
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Start Inverse Trigonometric Functions - Choose Your ModeAbout Inverse Trigonometric Functions - Class 12 Mathematics
Inverse Trigonometric Functions is the second chapter of Class 12 Mathematics. Since trigonometric functions are not one-one over their natural domains, we restrict their domains to specific intervals where they become bijective, and then define their inverses. The restricted sin function on [−π/2, π/2] gives sin⁻¹, the restricted cos on [0, π] gives cos⁻¹, and the restricted tan on (−π/2, π/2) gives tan⁻¹.
The chapter establishes key properties: sin⁻¹(x) + cos⁻¹(x) = π/2 for all x in [−1, 1], and similar complementary identities for the other pairs. The addition formula tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1−xy)) (when xy < 1) is particularly important for proving identities. The double-angle formulas like 2tan⁻¹(x) = sin⁻¹(2x/(1+x²)) connect different inverse functions.
This chapter is important for both the board exam (4 marks) and competitive exams like JEE Main, where inverse trigonometric identities are frequently tested. The principal value concept is fundamental - students must understand that sin⁻¹(sin(x)) = x only when x lies in the principal branch, and learn to handle cases when it does not.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 2?
Chapter 2 Inverse Trigonometric Functions has 2 exercises plus a Miscellaneous Exercise. Exercise 2.1 has 14 questions on principal values, Exercise 2.2 has 21 questions on properties and proofs, and the Miscellaneous Exercise has 17 questions combining all concepts.
How many marks does Inverse Trigonometric Functions carry in CBSE Class 12?
Inverse Trigonometric Functions carries about 4 marks in the CBSE Class 12 board exam 2027. Combined with Chapter 1 (Relations and Functions), the unit total is 8 marks. Expect 1 MCQ (1 mark) and 1 short answer (2-3 marks) from this chapter.
What is the principal value of sin⁻¹(x)?
The principal value of sin⁻¹(x) is the unique value y in the range [−π/2, π/2] such that sin(y) = x. For example, sin⁻¹(1/2) = π/6 (not 5π/6, which also satisfies sin(y) = 1/2 but is outside the principal range).
What is the difference between sin⁻¹(x) and (sin x)⁻¹?
sin⁻¹(x) is the inverse trigonometric function - it gives the angle whose sine is x. (sin x)⁻¹ = 1/sin(x) = cosec(x), which is the reciprocal of sine. These are completely different. sin⁻¹(1/2) = π/6, but (sin(π/6))⁻¹ = 1/(1/2) = 2.
Why is sin⁻¹(sin(2π/3)) not equal to 2π/3?
Because 2π/3 is outside the principal value range [−π/2, π/2] of sin⁻¹. The function sin⁻¹(sin(x)) = x only when x ∈ [−π/2, π/2]. For 2π/3, we compute: sin(2π/3) = sin(π − π/3) = sin(π/3) = √3/2. Then sin⁻¹(√3/2) = π/3.
Is NCERT enough for Inverse Trigonometric Functions?
Yes, NCERT is sufficient for the board exam. Focus on Exercise 2.1 (principal values) and the identity-based proofs in Exercise 2.2 and the Miscellaneous Exercise. The board questions follow NCERT patterns closely. For JEE, also solve exemplar problems.
What are the most important identities in this chapter?
The three most important identities are: (1) sin⁻¹(x) + cos⁻¹(x) = π/2, (2) tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1−xy)) when xy < 1, and (3) 2tan⁻¹(x) = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²)). These three identities cover almost every board exam question from this chapter.
How do I remember the domain and range of all six inverse trig functions?
Remember the three complementary pairs: sin⁻¹ and cosec⁻¹ have range [−π/2, π/2], cos⁻¹ and sec⁻¹ have range [0, π], tan⁻¹ has range (−π/2, π/2) and cot⁻¹ has range (0, π). For domains: sin⁻¹ and cos⁻¹ take [−1, 1], while sec⁻¹ and cosec⁻¹ take R − (−1, 1). tan⁻¹ and cot⁻¹ take all of R.