Class 12 Mathematics CBSE Format

NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

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NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions provides students with step-by-step answers and conceptual clarity for every problem in the official CBSE syllabus. Inverse trigonometric functions (often abbreviated as ITF) form an essential pillar of senior secondary mathematics. Not only does this chapter carry direct marks in Unit 1 (Relations and Functions), but mastering its definitions, restrictions, and identities is also vital for excelling in Calculus, including Continuity, Differentiability, Integrals, and Differential Equations.

In this detailed guide, you will find complete, curriculum-aligned NCERT Mathematics Class 12 solutions with rigorous algebraic proofs, principal value evaluations, domain-range tables, and common pitfalls to avoid during your CBSE board examinations.

Chapter Overview: Inverse Trigonometric Functions

In standard trigonometry, trigonometric functions like sine, cosine, and tangent map real angles (in radians or degrees) to real numbers. Because basic trigonometric functions are periodic, they are many-to-one over their natural domains. However, an inverse function exists if and only if the original function is bijective (both one-to-one/injective and onto/surjective).

To define inverse trigonometric functions, mathematicians restrict the domain of each trigonometric function to an interval where it is monotonic and bijective. The resulting range of the inverse function is known as the Principal Value Branch (PVB). Chapter 2 is divided into two primary exercises and one miscellaneous exercise in the NCERT textbook:

  • Exercise 2.1: Determination of principal values of standard inverse trigonometric expressions using interval definitions.
  • Exercise 2.2: Simplification of complex algebraic-trigonometric expressions, substitution methods, and self-inversion evaluations.
  • Miscellaneous Exercise: Comprehensive multi-concept problems, identities, and solving trigonometric equations.

Important Formulas, Domains, and Principal Value Branches

Memorizing the correct domain and principal value range for each inverse trigonometric function is critical for solving CBSE board exam questions without committing boundary errors.

Function Symbol Domain (Valid Input x) Principal Value Branch / Range (Output y)
Arcsine y = sin−1(x) [−1, 1] [−π/2, π/2]
Arccosine y = cos−1(x) [−1, 1] [0, π]
Arctangent y = tan−1(x) R (All Real Numbers) (−π/2, π/2)
Arccosecant y = cosec−1(x) R − (−1, 1) or |x| ≥ 1 [−π/2, π/2] − {0}
Arcsecant y = sec−1(x) R − (−1, 1) or |x| ≥ 1 [0, π] − {π/2}
Arccotangent y = cot−1(x) R (All Real Numbers) (0, π)

Fundamental Properties and Reduction Identities

1. Negative Argument Identities:

  • Group A (Odd-Function Symmetry):
    • sin−1(−x) = −sin−1(x), for x ∈ [−1, 1]
    • tan−1(−x) = −tan−1(x), for x ∈ R
    • cosec−1(−x) = −cosec−1(x), for |x| ≥ 1
  • Group B (Supplementary Angle Symmetry):
    • cos−1(−x) = π − cos−1(x), for x ∈ [−1, 1]
    • sec−1(−x) = π − sec−1(x), for |x| ≥ 1
    • cot−1(−x) = π − cot−1(x), for x ∈ R

2. Complementary Angle Sums:

  • sin−1(x) + cos−1(x) = π/2, for x ∈ [−1, 1]
  • tan−1(x) + cot−1(x) = π/2, for x ∈ R
  • sec−1(x) + cosec−1(x) = π/2, for |x| ≥ 1

3. Self-Inversion Properties:

  • sin−1(sin θ) = θ only if θ ∈ [−π/2, π/2]
  • cos−1(cos θ) = θ only if θ ∈ [0, π]
  • tan−1(tan θ) = θ only if θ ∈ (−π/2, π/2)

NCERT Solutions Class 12 Mathematics Chapter 2: Step-by-Step Exercise Solutions

Exercise 2.1: Principal Value Branch Calculations

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

Given: The expression sin−1(−1/2).

To Find: The principal value in the range [−π/2, π/2].

Step-by-Step Solution:

  1. Let y = sin−1(−1/2).
  2. By definition, sin(y) = −1/2.
  3. We know that sin(π/6) = 1/2. Therefore, −sin(π/6) = sin(−π/6).
  4. So, sin(y) = sin(−π/6).
  5. Since −π/6 lies strictly within the principal value branch [−π/2, π/2], we have y = −π/6.

Answer: −π/6

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

Given: The expression cos−1(−1/2).

To Find: The principal value in the range [0, π].

Step-by-Step Solution:

  1. Let y = cos−1(−1/2).
  2. This implies cos(y) = −1/2.
  3. We know that cos(π/3) = 1/2.
  4. Using the identity cos(π − θ) = −cos(θ), we get:
    cos(y) = −cos(π/3) = cos(π − π/3) = cos(2π/3).
  5. Since 2π/3 ∈ [0, π], the principal value is 2π/3.

Answer: 2π/3

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

Given: E = tan−1(1) + cos−1(−1/2) + sin−1(−1/2).

To Evaluate: The sum of the principal values.

Method 1 (Direct Evaluation):

  1. tan−1(1) = π/4 (since tan(π/4) = 1 and π/4 ∈ (−π/2, π/2)).
  2. cos−1(−1/2) = π − cos−1(1/2) = π − π/3 = 2π/3.
  3. sin−1(−1/2) = −sin−1(1/2) = −π/6.
  4. Sum = π/4 + 2π/3 − π/6 = π/4 + (4π − π)/6 = π/4 + 3π/6 = π/4 + π/2 = 3π/4.

Method 2 (Using Complementary Identity):

  1. Recall that cos−1(x) + sin−1(x) = π/2 for all x ∈ [−1, 1].
  2. Here, cos−1(−1/2) + sin−1(−1/2) = π/2.
  3. Therefore, E = tan−1(1) + π/2 = π/4 + π/2 = 3π/4.

Answer: 3π/4

Question 14: Find the value of tan−1(√3) − sec−1(−2)

Step-by-Step Solution:

  1. Let tan−1(√3) = α. Since tan(π/3) = √3 and π/3 ∈ (−π/2, π/2), we have α = π/3.
  2. Let sec−1(−2) = β. Using sec−1(−x) = π − sec−1(x):
    β = π − sec−1(2) = π − π/3 = 2π/3.
  3. Now compute the difference:
    tan−1(√3) − sec−1(−2) = π/3 − 2π/3 = −π/3.

Answer: −π/3

Exercise 2.2: Simplification and Evaluation of Inverse Trigonometric Functions

Question 1: Prove that 3 sin−1(x) = sin−1(3x − 4x3), x ∈ [−1/2, 1/2]

Proof:

  1. Let sin−1(x) = θ ⇒ x = sin(θ).
  2. Since x ∈ [−1/2, 1/2], we have θ ∈ [−π/6, π/6].
  3. Multiplying by 3 gives 3θ ∈ [−π/2, π/2], which falls cleanly within the principal value branch of arcsine.
  4. We know the standard triple-angle formula: sin(3θ) = 3 sin(θ) − 4 sin3(θ).
  5. Substituting x = sin(θ):
    sin(3θ) = 3x − 4x3 ⇒ 3θ = sin−1(3x − 4x3).
  6. Replacing θ with sin−1(x):
    3 sin−1(x) = sin−1(3x − 4x3).

Result: Hence Proved.

Question 5: Write in simplest form: tan−1[(√(1 + x2) − 1) / x], x ≠ 0

Step-by-Step Solution:

  1. Substitute x = tan(θ) ⇒ θ = tan−1(x), where θ ∈ (−π/2, π/2) − {0}.
  2. The expression inside the brackets becomes:
    [√(1 + tan2θ) − 1] / tan(θ) = [sec(θ) − 1] / tan(θ).
  3. Convert to sine and cosine:
    [ (1/cos(θ)) − 1 ] / [ sin(θ)/cos(θ) ] = (1 − cos(θ)) / sin(θ).
  4. Apply half-angle identities:
    1 − cos(θ) = 2 sin2(θ/2) and sin(θ) = 2 sin(θ/2) cos(θ/2).
  5. Substitute these back:
    [ 2 sin2(θ/2) ] / [ 2 sin(θ/2) cos(θ/2) ] = sin(θ/2) / cos(θ/2) = tan(θ/2).
  6. Now evaluate the full expression:
    tan−1(tan(θ/2)) = θ/2 (since θ ∈ (−π/2, π/2) ⇒ θ/2 ∈ (−π/4, π/4)).
  7. Substitute back θ = tan−1(x):
    = (1/2) tan−1(x).

Answer: (1/2) tan−1(x)

Question 8: Write in simplest form: tan−1[(cos x − sin x) / (cos x + sin x)], −π/4 < x < 3π/4

Step-by-Step Solution:

  1. Divide both the numerator and denominator by cos(x):
    (cos x − sin x) / (cos x + sin x) = (1 − tan x) / (1 + tan x).
  2. Use the fact that tan(π/4) = 1:
    = [tan(π/4) − tan x] / [1 + tan(π/4) tan x].
  3. Recall the subtraction formula tan(A − B) = (tan A − tan B) / (1 + tan A tan B):
    = tan(π/4 − x).
  4. Now evaluate tan−1[tan(π/4 − x)]:
    Given −π/4 < x < 3π/4 ⇒ −3π/4 < −x < π/4 ⇒ −π/2 < π/4 − x < π/2.
  5. Since (π/4 − x) lies strictly within the principal branch (−π/2, π/2), we get:
    tan−1[tan(π/4 − x)] = π/4 − x.

Answer: π/4 − x

Question 9: Find the value of sin−1(sin(2π/3))

Step-by-Step Solution:

  1. Common Mistake: Writing sin−1(sin(2π/3)) = 2π/3. This is incorrect because 2π/3 ∉ [−π/2, π/2].
  2. Rewrite 2π/3 in the second quadrant: 2π/3 = π − π/3.
  3. Using sin(π − θ) = sin(θ), we have sin(2π/3) = sin(π − π/3) = sin(π/3).
  4. Now substitute back:
    sin−1(sin(2π/3)) = sin−1(sin(π/3)).
  5. Since π/3 ∈ [−π/2, π/2], the value is π/3.

Answer: π/3

Question 10: Find the value of tan−1(tan(3π/4))

Step-by-Step Solution:

  1. Check the angle: 3π/4 = 135°, which is outside the range (−π/2, π/2).
  2. Express 3π/4 as π − π/4.
  3. Using tan(π − θ) = −tan(θ), we get:
    tan(3π/4) = tan(π − π/4) = −tan(π/4) = tan(−π/4).
  4. Now calculate:
    tan−1(tan(−π/4)) = −π/4, which lies in (−π/2, π/2).

Answer: −π/4

Miscellaneous Exercise on Chapter 2: Key NCERT Problems

Question: Find the value of cos−1(cos(13π/6))

Step-by-Step Solution:

  1. 13π/6 ∉ [0, π], because 13π/6 = 2π + π/6.
  2. Using periodicity: cos(2π + θ) = cos(θ), so cos(13π/6) = cos(2π + π/6) = cos(π/6).
  3. Now, cos−1(cos(13π/6)) = cos−1(cos(π/6)).
  4. Since π/6 ∈ [0, π], the result is π/6.

Answer: π/6

Question: Prove that sin−1(3/5) − sin−1(8/17) = cos−1(84/85)

Step-by-Step Solution:

  1. Let A = sin−1(3/5) ⇒ sin(A) = 3/5.
    Then cos(A) = √(1 − (3/5)2) = √(16/25) = 4/5.
  2. Let B = sin−1(8/17) ⇒ sin(B) = 8/17.
    Then cos(B) = √(1 − (8/17)2) = √(225/289) = 15/17.
  3. We need the result in terms of arccosine. Use the subtraction formula for cosine:
    cos(A − B) = cos(A) cos(B) + sin(A) sin(B).
  4. Substitute values:
    cos(A − B) = (4/5) × (15/17) + (3/5) × (8/17)
    cos(A − B) = (60/85) + (24/85) = 84/85.
  5. Taking arccosine on both sides (since 84/85 > 0, A − B ∈ [0, π]):
    A − B = cos−1(84/85).
  6. Substitute back A and B:
    sin−1(3/5) − sin−1(8/17) = cos−1(84/85).

Result: Hence Proved.

Question: Solve for x: 2 tan−1(cos x) = tan−1(2 cosec x)

Step-by-Step Solution:

  1. Recall the identity: 2 tan−1(y) = tan−1(2y / (1 − y2)).
  2. Apply this to the LHS with y = cos(x):
    2 tan−1(cos x) = tan−1[ 2 cos(x) / (1 − cos2(x)) ] = tan−1[ 2 cos(x) / sin2(x) ].
  3. Equate LHS and RHS:
    tan−1[ 2 cos(x) / sin2(x) ] = tan−1[ 2 cosec(x) ].
  4. Taking tangent of both sides:
    2 cos(x) / sin2(x) = 2 / sin(x).
  5. Assuming sin(x) ≠ 0 (since cosec(x) is defined):
    cos(x) / sin(x) = 1 ⇒ cot(x) = 1 ⇒ tan(x) = 1.
  6. For x in the fundamental interval [0, π], x = π/4.

Answer: x = π/4

Common Mistakes and Tips for CBSE Class 12 Students

Inverse trigonometry is straightforward once concepts are clear, but minor oversights can cost valuable marks in board examinations. Keep these critical pointers in mind:

  • Confusing Notation: Never write sin−1(x) as (sin x)−1. The former represents the inverse angle, while (sin x)−1 = 1 / sin(x) = cosec(x).
  • Ignoring Principal Value Intervals: Do not automatically equate f−1(f(θ)) = θ. Always check if θ lies in the principal value branch. If not, use quadrant reduction formulas (like π − θ, π + θ, or 2π − θ) to bring it into the valid range.
  • Negative Arguments for Cosine, Secant, and Cotangent: Remember that cos−1(−x) = π − cos−1(x), NOT −cos−1(x). This is because the range of cos−1(x) is [0, π], which cannot contain negative angles.
  • Checking Extraneous Solutions: When solving equations involving inverse trigonometric functions, always substitute the obtained values of x back into the original equation to ensure they satisfy the domain conditions (|x| ≤ 1 for sin−1 and cos−1).

Board Exam Relevance and Mark Distribution

In the CBSE Class 12 Mathematics blueprint, Chapter 2 (Inverse Trigonometric Functions) contributes directly to Unit 1: Relations and Functions (total unit weightage: 10 marks). Typically, Chapter 2 accounts for 4 to 6 direct marks across various question formats:

  • 1-Mark MCQs / Assertion-Reason: Standard questions testing knowledge of the principal value branch, domain ranges, or direct simplifications like sin(cos−1(3/5)).
  • 2-Mark Very Short Answer (VSA): Direct evaluation of expressions with multiple terms, e.g., tan−1(1) + cos−1(−1/2).
  • 3-Mark / 4-Mark Short Answer or Case-Based: Simplification of algebraic rational expressions via trigonometric substitution, or solving inverse trigonometric equations.
  • Calculus Foundation: Over 25% of questions in Chapter 5 (Continuity & Differentiability) and Chapter 7 (Integrals) require substitution techniques learned in Chapter 2.

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