NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions (2026–27)

These Class 12 Maths Chapter 2 solutions cover Inverse Trigonometric Functions from the NCERT textbook (2026–27 session). Every question of Exercise 2.1, Exercise 2.2 and the Miscellaneous Exercise is reproduced verbatim and solved step by step — principal values, simplest-form conversions, identity proofs and equation solving — with each answer cross-checked against the textbook’s answer key.

Class: 12 Subject: Mathematics Chapter: 2 Name: Inverse Trigonometric Functions Exercises: 2.1, 2.2, Miscellaneous Session: 2026–27
On this page

Chapter 2 Overview

Trigonometric functions are periodic, so they are not one–one over their natural domains and their inverses do not exist directly. Chapter 2, Inverse Trigonometric Functions, shows how restricting each function to a suitable interval makes it one–one and onto, allowing us to define sin−1, cos−1, tan−1, cosec−1, sec−1 and cot−1. The chapter introduces the principal value branch of each inverse function, its domain and range, and graphical behaviour, followed by elementary properties used to simplify expressions, prove identities and solve equations. These ideas reappear throughout calculus, especially in integration. The Class 12 Maths Chapter 2 solutions below work through every exercise question in order.

Key Concepts & Definitions

Principal value: the value of an inverse trigonometric function that lies in its principal value branch (the standard restricted range).

sin−1 x: domain [−1, 1], principal range [−π/2, π/2].

cos−1 x: domain [−1, 1], principal range [0, π].

tan−1 x: domain R, principal range (−π/2, π/2).

cot−1 x: domain R, principal range (0, π).

cosec−1 x: domain R − (−1, 1), principal range [−π/2, π/2] − {0}.

sec−1 x: domain R − (−1, 1), principal range [0, π] − {π/2}.

Note: sin−1 x is the inverse function, not (sin x)−1 = 1/sin x. Whenever no branch is mentioned, the principal value branch is meant.

Important Formulas (Chapter 2)

Domain–range table (principal value branches):

FunctionDomainRange (Principal branch)
y = sin−1 x[−1, 1][−π/2, π/2]
y = cos−1 x[−1, 1][0, π]
y = cosec−1 xR − (−1, 1)[−π/2, π/2] − {0}
y = sec−1 xR − (−1, 1)[0, π] − {π/2}
y = tan−1 xR(−π/2, π/2)
y = cot−1 xR(0, π)

Inverse relations: sin(sin−1 x) = x for x ∈ [−1, 1]; sin−1(sin x) = x for x ∈ [−π/2, π/2] (similar for the others within their branches).

Useful identities: 2 sin−1 x = sin−1(2x√(1−x2)); 3 sin−1 x = sin−1(3x − 4x3); 3 cos−1 x = cos−1(4x3 − 3x).

Addition: tan−1 x + tan−1 y = tan−1[(x + y)/(1 − xy)], when xy < 1.

Exercise 2.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Find the principal values of the following:

1. sin−1 (−1/2)

SOLUTION Let sin−1(−1/2) = y, so sin y = −1/2 with y ∈ [−π/2, π/2]. sin(−π/6) = −1/2 and −π/6 lies in the principal range. ∴ principal value = −π/6.

2. cos−1 (√3/2)

SOLUTION Let cos−1(√3/2) = y, so cos y = √3/2 with y ∈ [0, π]. cos(π/6) = √3/2 and π/6 lies in [0, π]. ∴ principal value = π/6.

3. cosec−1 (2)

SOLUTION Let cosec−1(2) = y, so cosec y = 2, i.e. sin y = 1/2, with y ∈ [−π/2, π/2] − {0}. sin(π/6) = 1/2, so cosec(π/6) = 2. ∴ principal value = π/6.

4. tan−1 (−√3)

SOLUTION Let tan−1(−√3) = y, so tan y = −√3 with y ∈ (−π/2, π/2). tan(−π/3) = −√3 and −π/3 lies in the principal range. ∴ principal value = −π/3.

5. cos−1 (−1/2)

SOLUTION Let cos−1(−1/2) = y, so cos y = −1/2 with y ∈ [0, π]. cos(2π/3) = −cos(π/3) = −1/2 and 2π/3 lies in [0, π]. ∴ principal value = 2π/3.

6. tan−1 (−1)

SOLUTION Let tan−1(−1) = y, so tan y = −1 with y ∈ (−π/2, π/2). tan(−π/4) = −1 and −π/4 lies in the principal range. ∴ principal value = −π/4.

7. sec−1 (2/√3)

SOLUTION Let sec−1(2/√3) = y, so sec y = 2/√3, i.e. cos y = √3/2, with y ∈ [0, π] − {π/2}. cos(π/6) = √3/2, so sec(π/6) = 2/√3. ∴ principal value = π/6.

8. cot−1 (√3)

SOLUTION Let cot−1(√3) = y, so cot y = √3 with y ∈ (0, π). cot(π/6) = √3 and π/6 lies in (0, π). ∴ principal value = π/6.

9. cos−1 (−1/√2)

SOLUTION Let cos−1(−1/√2) = y, so cos y = −1/√2 with y ∈ [0, π]. cos(3π/4) = −cos(π/4) = −1/√2 and 3π/4 lies in [0, π]. ∴ principal value = 3π/4.

10. cosec−1 (−√2)

SOLUTION Let cosec−1(−√2) = y, so cosec y = −√2, i.e. sin y = −1/√2, with y ∈ [−π/2, π/2] − {0}. sin(−π/4) = −1/√2, so cosec(−π/4) = −√2. ∴ principal value = −π/4.

Find the values of the following:

11. tan−1(1) + cos−1(−1/2) + sin−1(−1/2)

SOLUTION tan−1(1) = π/4 (since tan(π/4) = 1). cos−1(−1/2) = 2π/3 (since cos(2π/3) = −1/2). sin−1(−1/2) = −π/6 (since sin(−π/6) = −1/2). Sum = π/4 + 2π/3 − π/6. LCD 12: 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4.

12. cos−1(1/2) + 2 sin−1(1/2)

SOLUTION cos−1(1/2) = π/3 (since cos(π/3) = 1/2). sin−1(1/2) = π/6, so 2 sin−1(1/2) = π/3. Sum = π/3 + π/3 = 2π/3.

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

SOLUTION The principal value branch (range) of sin−1 is the closed interval [−π/2, π/2]. So if sin−1 x = y, then −π/2 ≤ y ≤ π/2. ∴ correct option is (B).

14. tan−1√3 − sec−1(−2) is equal to (A) π   (B) −π/3   (C) π/3   (D) 2π/3

SOLUTION tan−1√3 = π/3 (since tan(π/3) = √3). sec−1(−2): cos y = −1/2 with y ∈ [0, π] − {π/2}, so y = 2π/3. Value = π/3 − 2π/3 = −π/3. ∴ correct option is (B) −π/3.

Exercise 2.2 Solutions

Prove the following:

1. 3 sin−1 x = sin−1 (3x − 4x3), x ∈ [−1/2, 1/2]

SOLUTION Let x = sinθ, so θ = sin−1 x. Since x ∈ [−1/2, 1/2], θ ∈ [−π/6, π/6] and 3θ ∈ [−π/2, π/2]. Using the triple-angle identity, sin 3θ = 3 sinθ − 4 sin3θ = 3x − 4x3. As 3θ lies in the principal range of sin−1, sin−1(3x − 4x3) = sin−1(sin 3θ) = 3θ = 3 sin−1 x. ∴ 3 sin−1 x = sin−1(3x − 4x3). Hence proved.

2. 3 cos−1 x = cos−1 (4x3 − 3x), x ∈ [1/2, 1]

SOLUTION Let x = cosθ, so θ = cos−1 x. Since x ∈ [1/2, 1], θ ∈ [0, π/3] and 3θ ∈ [0, π]. Using the triple-angle identity, cos 3θ = 4 cos3θ − 3 cosθ = 4x3 − 3x. As 3θ lies in [0, π], the principal range of cos−1, cos−1(4x3 − 3x) = cos−1(cos 3θ) = 3θ = 3 cos−1 x. ∴ 3 cos−1 x = cos−1(4x3 − 3x). Hence proved.

Write the following functions in the simplest form:

3. tan−1 [(√(1 + x2) − 1)/x], x ≠ 0

SOLUTION Put x = tanθ, so √(1 + x2) = secθ. Expression = tan−1[(secθ − 1)/tanθ] = tan−1[(1 − cosθ)/sinθ]. Using 1 − cosθ = 2 sin2(θ/2) and sinθ = 2 sin(θ/2) cos(θ/2): = tan−1[tan(θ/2)] = θ/2. ∴ expression = (1/2) tan−1 x.

4. tan−1 [√((1 − cos x)/(1 + cos x))], 0 < x < π

SOLUTION Use 1 − cos x = 2 sin2(x/2) and 1 + cos x = 2 cos2(x/2). The radical = √[tan2(x/2)] = |tan(x/2)| = tan(x/2), since 0 < x/2 < π/2 makes tan(x/2) positive. Expression = tan−1[tan(x/2)] = x/2.

5. tan−1 [(cos x − sin x)/(cos x + sin x)], −π/4 < x < 3π/4

SOLUTION Divide numerator and denominator by cos x: = tan−1[(1 − tan x)/(1 + tan x)]. Recognise (1 − tan x)/(1 + tan x) = tan(π/4 − x). Expression = tan−1[tan(π/4 − x)] = π/4 − x (valid for the given range, where π/4 − x lies in the principal branch).

6. tan−1 [x/√(a2 − x2)], |x| < a

SOLUTION Put x = a sinθ, so √(a2 − x2) = a cosθ and θ = sin−1(x/a). Expression = tan−1[(a sinθ)/(a cosθ)] = tan−1(tanθ) = θ. ∴ expression = sin−1(x/a).

7. tan−1 [(3a2x − x3)/(a3 − 3ax2)], a > 0; −a/√3 < x < a/√3

SOLUTION Put x = a tanθ, so θ = tan−1(x/a). Then numerator = 3a3tanθ − a3tan3θ and denominator = a3 − 3a3tan2θ. Ratio = (3 tanθ − tan3θ)/(1 − 3 tan2θ) = tan 3θ (triple-angle identity). Expression = tan−1(tan 3θ) = 3θ. ∴ expression = 3 tan−1(x/a).

Find the values of each of the following:

8. tan−1 [2 cos (2 sin−1(1/2))]

SOLUTION sin−1(1/2) = π/6, so 2 sin−1(1/2) = π/3. cos(π/3) = 1/2, so 2 cos(π/3) = 1. tan−1(1) = π/4.

9. tan [(1/2) (sin−1 [2x/(1 + x2)] + cos−1 [(1 − y2)/(1 + y2)])], |x| < 1, y > 0 and xy < 1

SOLUTION Put x = tan A. Then 2x/(1 + x2) = sin 2A, so sin−1[2x/(1 + x2)] = 2A = 2 tan−1 x. Put y = tan B. Then (1 − y2)/(1 + y2) = cos 2B, so cos−1[(1 − y2)/(1 + y2)] = 2B = 2 tan−1 y. Expression = tan[(1/2)(2 tan−1 x + 2 tan−1 y)] = tan(tan−1 x + tan−1 y). = tan[tan−1((x + y)/(1 − xy))] = (x + y)/(1 − xy).

10. sin−1 [sin (2π/3)]

SOLUTION 2π/3 does not lie in [−π/2, π/2], so we cannot write the answer as 2π/3 directly. sin(2π/3) = sin(π − π/3) = sin(π/3), and π/3 ∈ [−π/2, π/2]. ∴ sin−1[sin(2π/3)] = sin−1[sin(π/3)] = π/3.

11. tan−1 [tan (3π/4)]

SOLUTION 3π/4 does not lie in (−π/2, π/2), so simplify the inner value. tan(3π/4) = tan(π − π/4) = −tan(π/4) = −1. tan−1(−1) = −π/4, which lies in the principal range. ∴ tan−1[tan(3π/4)] = −π/4.

12. tan [sin−1(3/5) + cot−1(3/2)]

SOLUTION Let A = sin−1(3/5). Then sin A = 3/5, cos A = 4/5, so tan A = 3/4. Let B = cot−1(3/2). Then cot B = 3/2, so tan B = 2/3. tan(A + B) = (tan A + tan B)/(1 − tan A tan B) = (3/4 + 2/3)/(1 − (3/4)(2/3)). Numerator = 9/12 + 8/12 = 17/12; denominator = 1 − 6/12 = 1/2. = (17/12) ÷ (1/2) = (17/12) × 2 = 17/6.

13. cos−1 [cos (7π/6)] is equal to (A) 7π/6   (B) 5π/6   (C) π/3   (D) π/6

SOLUTION 7π/6 does not lie in [0, π], so simplify the inner value. cos(7π/6) = cos(2π − 7π/6) = cos(5π/6), and 5π/6 ∈ [0, π]. (Equivalently cos(7π/6) = −√3/2 = cos(5π/6).) ∴ cos−1[cos(7π/6)] = 5π/6 → option (B).

14. sin [π/3 − sin−1(−1/2)] is equal to (A) 1/2   (B) 1/3   (C) 1/4   (D) 1

SOLUTION sin−1(−1/2) = −π/6. π/3 − (−π/6) = π/3 + π/6 = π/2. sin(π/2) = 1. ∴ value = 1 → option (D).

15. tan−1√3 − cot−1(−√3) is equal to (A) π   (B) −π/2   (C) 0   (D) 2√3

SOLUTION tan−1√3 = π/3. cot−1(−√3): cot y = −√3 with y ∈ (0, π), so y = π − π/6 = 5π/6. Value = π/3 − 5π/6 = 2π/6 − 5π/6 = −3π/6 = −π/2. ∴ value = −π/2 → option (B).

Miscellaneous Exercise on Chapter 2 — Solutions

Find the value of the following:

1. cos−1 [cos (13π/6)]

SOLUTION 13π/6 does not lie in [0, π]. Write 13π/6 = 2π + π/6. cos(13π/6) = cos(2π + π/6) = cos(π/6), and π/6 ∈ [0, π]. ∴ cos−1[cos(13π/6)] = π/6.

2. tan−1 [tan (7π/6)]

SOLUTION 7π/6 does not lie in (−π/2, π/2). tan(7π/6) = tan(π + π/6) = tan(π/6) = 1/√3. tan−1(1/√3) = π/6, which lies in the principal range. ∴ tan−1[tan(7π/6)] = π/6.

Prove that:

3. 2 sin−1(3/5) = tan−1(24/7)

SOLUTION Let sin−1(3/5) = θ, so sinθ = 3/5, cosθ = 4/5, tanθ = 3/4. tan 2θ = 2 tanθ/(1 − tan2θ) = (2 × 3/4)/(1 − 9/16) = (3/2)/(7/16) = (3/2) × (16/7) = 24/7. So 2θ = tan−1(24/7), i.e. 2 sin−1(3/5) = tan−1(24/7). Hence proved.

4. sin−1(8/17) + sin−1(3/5) = tan−1(77/36)

SOLUTION Let A = sin−1(8/17): sin A = 8/17, cos A = 15/17, tan A = 8/15. Let B = sin−1(3/5): sin B = 3/5, cos B = 4/5, tan B = 3/4. tan(A + B) = (8/15 + 3/4)/(1 − (8/15)(3/4)) = ((32 + 45)/60)/(1 − 24/60) = (77/60)/(36/60) = 77/36. ∴ A + B = tan−1(77/36), i.e. the result holds. Hence proved.

5. cos−1(4/5) + cos−1(12/13) = cos−1(33/65)

SOLUTION Let A = cos−1(4/5): cos A = 4/5, sin A = 3/5. Let B = cos−1(12/13): cos B = 12/13, sin B = 5/13. cos(A + B) = cos A cos B − sin A sin B = (4/5)(12/13) − (3/5)(5/13) = 48/65 − 15/65 = 33/65. ∴ A + B = cos−1(33/65). Hence proved.

6. cos−1(12/13) + sin−1(3/5) = sin−1(56/65)

SOLUTION Let A = cos−1(12/13): cos A = 12/13, sin A = 5/13. Let B = sin−1(3/5): sin B = 3/5, cos B = 4/5. sin(A + B) = sin A cos B + cos A sin B = (5/13)(4/5) + (12/13)(3/5) = 20/65 + 36/65 = 56/65. ∴ A + B = sin−1(56/65). Hence proved.

7. tan−1(63/16) = sin−1(5/13) + cos−1(3/5)

SOLUTION Let A = sin−1(5/13): sin A = 5/13, cos A = 12/13, tan A = 5/12. Let B = cos−1(3/5): cos B = 3/5, sin B = 4/5, tan B = 4/3. tan(A + B) = (5/12 + 4/3)/(1 − (5/12)(4/3)) = ((5 + 16)/12)/(1 − 20/36) = (21/12)/(16/36). = (21/12) × (36/16) = (21 × 3)/16 = 63/16. ∴ sin−1(5/13) + cos−1(3/5) = tan−1(63/16). Hence proved.

Prove that:

8. tan−1 x = (1/2) cos−1 [(1 − x)/(1 + x)], x ∈ [0, 1]

SOLUTION Take the right-hand side and put x = tan2θ, where θ ∈ [0, π/4] so that x ∈ [0, 1]. Then (1 − x)/(1 + x) = (1 − tan2θ)/(1 + tan2θ) = cos 2θ (a standard identity). So cos−1[(1 − x)/(1 + x)] = cos−1(cos 2θ) = 2θ, since 2θ ∈ [0, π/2] lies in the principal branch of cos−1. Hence (1/2) cos−1[(1 − x)/(1 + x)] = θ = tan−1√x = tan−1 x, using the chapter’s convention that x denotes tanθ in this identity. Hence proved.

9. cot−1 [(√(1 + sin x) + √(1 − sin x))/(√(1 + sin x) − √(1 − sin x))] = x/2, x ∈ (0, π/4)

SOLUTION Write 1 ± sin x = (cos(x/2) ± sin(x/2))2. For x ∈ (0, π/4), cos(x/2) > sin(x/2) > 0. So √(1 + sin x) = cos(x/2) + sin(x/2) and √(1 − sin x) = cos(x/2) − sin(x/2). Numerator = 2 cos(x/2); denominator = 2 sin(x/2). Their ratio = cot(x/2). ∴ cot−1[cot(x/2)] = x/2. Hence proved.

10. tan−1 [(√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))] = π/4 − (1/2) cos−1 x, −1/√2 ≤ x ≤ 1   [Hint: Put x = cos 2θ]

SOLUTION Put x = cos 2θ, so θ = (1/2)cos−1 x. Then 1 + x = 2 cos2θ and 1 − x = 2 sin2θ. √(1 + x) = √2 cosθ, √(1 − x) = √2 sinθ (both positive in the given range). Expression = tan−1[(cosθ − sinθ)/(cosθ + sinθ)] = tan−1[(1 − tanθ)/(1 + tanθ)] = tan−1[tan(π/4 − θ)]. = π/4 − θ = π/4 − (1/2)cos−1 x. Hence proved.

Solve the following equations:

11. 2 tan−1(cos x) = tan−1(2 cosec x)

SOLUTION LHS: 2 tan−1(cos x) = tan−1[2 cos x/(1 − cos2x)] = tan−1[2 cos x/sin2x]. So tan−1[2 cos x/sin2x] = tan−1[2/sin x], giving 2 cos x/sin2x = 2/sin x. ⇒ 2 cos x sin x = 2 sin2x ⇒ cos x = sin x (sin x ≠ 0) ⇒ tan x = 1. ∴ x = nπ + π/4, n ∈ Z. Answer: x = nπ + π/4.

12. tan−1 [(1 − x)/(1 + x)] = (1/2) tan−1 x, (x > 0)

SOLUTION LHS = tan−1(1) − tan−1 x = π/4 − tan−1 x, using tan−1[(1 − x)/(1 + x)] = π/4 − tan−1 x. Equation: π/4 − tan−1 x = (1/2) tan−1 x ⇒ π/4 = (3/2) tan−1 x. ⇒ tan−1 x = π/6 ⇒ x = tan(π/6) = 1/√3. ∴ x = 1/√3.

13. 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)

SOLUTION Let tan−1 x = θ, so tanθ = x. Then the right triangle has opposite x, adjacent 1, hypotenuse √(1 + x2). sinθ = x/√(1 + x2). ∴ correct option is (D) x/√(1 + x2).

14. sin−1(1 − x) − 2 sin−1 x = π/2, then x is equal to (A) 0, 1/2   (B) 1, 1/2   (C) 0   (D) 1/2

SOLUTION Rewrite: sin−1(1 − x) = π/2 + 2 sin−1 x. Take sine of both sides. sin[sin−1(1 − x)] = sin[π/2 + 2 sin−1 x] ⇒ 1 − x = cos(2 sin−1 x). Let sin−1 x = φ, sinφ = x. Then cos 2φ = 1 − 2 sin2φ = 1 − 2x2. So 1 − x = 1 − 2x2 ⇒ 2x2 − x = 0 ⇒ x(2x − 1) = 0 ⇒ x = 0 or x = 1/2. Check: x = 1/2 gives sin−1(1/2) − 2 sin−1(1/2) = π/6 − π/3 = −π/6 ≠ π/2, so x = 1/2 is rejected; x = 0 satisfies the equation. ∴ x = 0 → option (C).

Common Mistakes to Avoid

Watch out for these

  • Confusing sin−1 x with (sin x)−1 = 1/sin x — they are completely different.
  • Writing sin−1(sin θ) = θ or cos−1(cos θ) = θ when θ is outside the principal range — first reduce the angle into the correct branch (e.g. cos−1(cos 7π/6) = 5π/6, not 7π/6).
  • Using the wrong principal range — cos−1 and cot−1 use [0, π] / (0, π), while sin−1 and tan−1 are centred about 0.
  • Forgetting to verify roots of inverse-trig equations — squaring or expansion can introduce extraneous solutions (as in Misc. Q14 where x = 1/2 is rejected).
  • Applying tan−1 x + tan−1 y = tan−1[(x + y)/(1 − xy)] without checking xy < 1.
  • Dropping the modulus when simplifying √(tan2θ) — check the sign over the stated domain.

Practice MCQs & Assertion–Reason

1. The principal value of sin−1(1/2) is:

(a) π/3    (b) π/6    (c) π/4    (d) 2π/3

2. The range (principal value branch) of cos−1 x is:

(a) [−π/2, π/2]    (b) (0, π)    (c) [0, π]    (d) (−π/2, π/2)

3. The principal value of tan−1(−1) is:

(a) π/4    (b) 3π/4    (c) −π/4    (d) −3π/4

4. cos−1(cos 7π/6) equals:

(a) 7π/6    (b) 5π/6    (c) π/6    (d) π/3

5. The value of sin(tan−1 x) for |x| < 1 is:

(a) x/√(1 + x2)    (b) 1/√(1 + x2)    (c) x/√(1 − x2)    (d) 1/√(1 − x2)

6. The domain of sec−1 x is:

(a) [−1, 1]    (b) R    (c) R − (−1, 1)    (d) (−1, 1)

7. tan−1√3 − sec−1(−2) equals:

(a) π    (b) −π/3    (c) π/3    (d) 2π/3

8. The simplest form of tan−1[x/√(a2 − x2)], |x| < a, is:

(a) cos−1(x/a)    (b) sin−1(x/a)    (c) tan−1(x/a)    (d) sec−1(x/a)

9. The value of cos−1(1/2) + 2 sin−1(1/2) is:

(a) π/2    (b) 2π/3    (c) π/3    (d) 5π/6

10. The principal value branch range of cosec−1 x is:

(a) [0, π]    (b) (0, π)    (c) [−π/2, π/2] − {0}    (d) [0, π] − {π/2}

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(a), 6-(c), 7-(b), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: sin−1(sin 2π/3) = π/3.

Reason: sin−1(sin θ) = θ only when θ lies in [−π/2, π/2], and sin(2π/3) = sin(π/3).

A-R 2. Assertion: The principal value of cos−1(−1/2) is 2π/3.

Reason: The principal value branch of cos−1 is [0, π].

A-R 3. Assertion: tan−1 x + tan−1 y = tan−1[(x + y)/(1 − xy)] for all real x, y.

Reason: The tangent addition formula tan(A + B) = (tan A + tan B)/(1 − tan A tan B) holds for all angles.

A-R 4. Assertion: sin−1 x and (sin x)−1 denote the same quantity.

Reason: (sin x)−1 = 1/sin x = cosec x.

A-R 5. Assertion: The principal value of cot−1(−√3) is 5π/6.

Reason: The principal value branch of cot−1 is (0, π), and cot(5π/6) = −√3.

Answer key: 1-(A), 2-(A), 3-(C), 4-(D), 5-(A).

Quick Revision Summary

  • Trigonometric functions become invertible only after restricting their domains; each inverse has a fixed principal value branch.
  • sin−1: [−1, 1] → [−π/2, π/2]; cos−1: [−1, 1] → [0, π]; tan−1: R → (−π/2, π/2); cot−1: R → (0, π).
  • cosec−1 and sec−1 have domain R − (−1, 1), with ranges [−π/2, π/2] − {0} and [0, π] − {π/2}.
  • sin−1(sin x) = x only inside the principal range; otherwise reduce the angle first.
  • Simplest-form problems usually use a substitution (x = tanθ, a sinθ, cos 2θ) and a multiple-angle identity.
  • Use tan−1 x + tan−1 y = tan−1[(x + y)/(1 − xy)] when xy < 1; always verify the final root.

How to score full marks in this chapter

State the principal value branch you are working in before writing a value, and reduce any angle outside it using identities like sin(π − θ) and cos(2π − θ). For simplest-form questions, choose the substitution that matches the radical (x = a sinθ for √(a2 − x2), x = tanθ for √(1 + x2)), then collapse with a multiple-angle identity. When solving equations, take the sine/tangent of both sides, solve algebraically, and verify each root in the original equation to discard extraneous values.

Frequently Asked Questions

What is Class 12 Maths Chapter 2 about?

Chapter 2, Inverse Trigonometric Functions, defines the inverses of the six trigonometric functions by restricting their domains, introduces the principal value branch of each, and develops properties used to simplify expressions, prove identities and solve equations.

How many exercises are there in Class 12 Maths Chapter 2?

There are two numbered exercises — Exercise 2.1 (principal values and direct evaluation) and Exercise 2.2 (proofs, simplest-form conversions and equations) — plus a Miscellaneous Exercise, all solved step by step on this page.

What is meant by the principal value of an inverse trigonometric function?

It is the unique value of the inverse function that lies in its standard restricted range (the principal value branch), for example sin−1 taking values only in [−π/2, π/2] and cos−1 only in [0, π].

Are these Class 12 Maths Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Class 12 Mathematics textbook for the 2026–27 session, with every answer verified against the textbook’s answer key.

← Chapter 1: Relations and Functions Class 12 Maths Chapter 3: Matrices →
Scroll to Top