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Chapter 7 · Class 12 Mathematics

Inverse Trigonometric Functions — Important Questions

33 questions With answers CBSE format

SUMMARY: This chapter focuses on the study of inverse trigonometric functions, their properties, and applications.
KEY TOPICS: definition of inverse trigonometric functions, principal value branches, domain and range, properties of inverse trigonometric functions, graphs of inverse trigonometric functions, solving equations involving inverse trigonometric functions, applications in calculus.

Previous chapter Integrals Next chapter Linear Programming

Multiple Choice Questions (MCQs) 5 questions

Q1 1 Mark

The principal value of sin⁻¹(1/2) is:

Aπ/6

Bπ/3

Cπ/4

Dπ/2

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Correct answer: Option 1 — π/6

Q2 1 Mark

The value of cos⁻¹(cos(7π/6)) is:

A7π/6

B5π/6

Cπ/6

D−π/6

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Correct answer: Option 2 — 5π/6

Q3 1 Mark

The principal value branch of sec⁻¹ x is:

A[0, π] − {π/2}

B[−π/2, π/2] − {0}

C[0, π/2)

D(−π/2, π/2)

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Correct answer: Option 1 — [0, π] − {π/2}

Q4 1 Mark

sin(2 sin⁻¹ x) equals:

A2x

B2x √(1 − x²)

Cx √(1 − x²)

D2x² − 1

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Correct answer: Option 2 — 2x √(1 − x²)

Q5 1 Mark

Domain of sin⁻¹(2x − 1) is:

A[−1, 1]

B[0, 1]

C[1, 2]

D[−1/2, 1/2]

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Correct answer: Option 2 — [0, 1]

Short Answer Questions 5 questions

Q6 3 Marks

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

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Principal value range of sin⁻¹ is [−π/2, π/2]. We need θ in this range such that sin θ = −1/2. Since sin(−π/6) = −1/2 and −π/6 ∈ [−π/2, π/2], the principal value is −π/6.

Q7 3 Marks

Prove that 2 tan⁻¹ x = sin⁻¹(2x / (1 + x²)) for |x| ≤ 1.

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Let tan⁻¹ x = θ, so tan θ = x, θ ∈ (−π/2, π/2). Then 2θ = 2 tan⁻¹ x. Now sin 2θ = 2 sin θ cos θ = 2 · (x / √(1+x²)) · (1 / √(1+x²)) = 2x / (1 + x²). So 2θ = sin⁻¹(2x / (1+x²)) provided 2θ ∈ [−π/2, π/2], i.e. |x| ≤ 1.

Q8 3 Marks

Evaluate cot⁻¹(cot(13π/8)).

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Range of cot⁻¹ is (0, π). Since 13π/8 = 2π − 3π/8 ∉ (0, π), reduce: cot(13π/8) = cot(13π/8 − π) = cot(5π/8). 5π/8 ∈ (0, π), so cot⁻¹(cot(13π/8)) = 5π/8.

Q9 3 Marks

Express tan⁻¹ x + tan⁻¹ y in terms of a single inverse function (state condition).

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For xy < 1: tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y) / (1 − xy)). For xy > 1, x > 0, y > 0: add π. For xy > 1, x < 0, y < 0: subtract π. The condition xy < 1 keeps the resulting angle in the principal value range of tan⁻¹.

Q10 3 Marks

Find the value of sin⁻¹(sin(3π/5)).

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3π/5 ∉ [−π/2, π/2]. Use sin(3π/5) = sin(π − 3π/5) = sin(2π/5). 2π/5 ∈ [−π/2, π/2]. Therefore sin⁻¹(sin(3π/5)) = 2π/5.

Long Answer Questions 6 questions

Q11 6 Marks

Show that 2 sin⁻¹ x = sin⁻¹(2 x √(1 − x²)) for x ∈ [−1/√2, 1/√2].

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Let sin⁻¹ x = θ, so sin θ = x, θ ∈ [−π/2, π/2]. We need 2θ ∈ [−π/2, π/2] which holds when θ ∈ [−π/4, π/4], i.e. x = sin θ ∈ [−1/√2, 1/√2]. Now sin 2θ = 2 sin θ cos θ = 2x · √(1 − x²) (since cos θ ≥ 0 in this range). Therefore 2θ = sin⁻¹(2x √(1 − x²)), i.e. 2 sin⁻¹ x = sin⁻¹(2x √(1 − x²)) on the stated domain.

Q12 6 Marks

Solve sin⁻¹ x + sin⁻¹(2x) = π/2 for x.

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From the equation, sin⁻¹(2x) = π/2 − sin⁻¹ x = cos⁻¹ x. So 2x = cos(cos⁻¹ x) = x — only if cos⁻¹ x = sin⁻¹(2x) which means 2x is the cosine: i.e. 2x = √(1 − x²). Squaring: 4x² = 1 − x² ⇒ 5x² = 1 ⇒ x = 1/√5 (taking positive root since 2x must be positive). Verify: sin⁻¹(1/√5) + sin⁻¹(2/√5) = sin⁻¹(1/√5) + cos⁻¹(1/√5) = π/2. ✓

Q13 6 Marks

Prove that tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = π.

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Use formula: tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 − xy)) + π if xy > 1 and x, y > 0. tan⁻¹(2) + tan⁻¹(3) = tan⁻¹((2+3)/(1−6)) + π = tan⁻¹(−1) + π = −π/4 + π = 3π/4. Add tan⁻¹(1) = π/4: total = π/4 + 3π/4 = π. Hence proved.

Q14 6 Marks

Solve for x: tan⁻¹(2x) + tan⁻¹(3x) = π/4.

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Use tan⁻¹ A + tan⁻¹ B = tan⁻¹((A+B)/(1−AB)) when AB < 1. Setting A = 2x, B = 3x: tan⁻¹((2x + 3x)/(1 − 6x²)) = π/4 ⇒ (5x)/(1 − 6x²) = tan(π/4) = 1 ⇒ 5x = 1 − 6x² ⇒ 6x² + 5x − 1 = 0. Factorise: (6x − 1)(x + 1) = 0 ⇒ x = 1/6 or x = −1. Check the AB < 1 constraint and the principal value range: x = 1/6 ⇒ 6x² = 1/6 < 1 ✓ and both inverse tangents are in (0, π/2). x = −1 ⇒ 6x² = 6 > 1, the additive formula needs a π adjustment, and gives 0 instead of π/4. Hence the only valid solution is x = 1/6.

Q15 6 Marks

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

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Let A = cos⁻¹(4/5) and B = cos⁻¹(12/13). Then cos A = 4/5, sin A = 3/5 (positive since A ∈ [0, π/2]). 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. Since A + B ∈ [0, π], A + B = cos⁻¹(33/65). Hence cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65). Proved.

Q16 6 Marks

Compare principal value branches of sin-1x and cos-1x with the help of a table.

Assertion–Reason Questions 5 questions

Q17 1 Mark

Assertion (A): sin⁻¹(1/2) = π/6.

Reason (R): π/6 lies in the principal value range [−π/2, π/2] and sin(π/6) = 1/2.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q18 1 Mark

Assertion (A): The principal value range of cos⁻¹ x is [0, π].

Reason (R): cos x is one-one on [0, π], so cos⁻¹ is well defined on this interval.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q19 1 Mark

Assertion (A): tan⁻¹(x) + cot⁻¹(x) = π/2 for all x ∈ R.

Reason (R): The two functions are complementary inverses by definition.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q20 1 Mark

Assertion (A): sin(2 sin⁻¹ x) = 2x √(1 − x²) for |x| ≤ 1.

Reason (R): This follows from the double-angle identity sin 2θ = 2 sin θ cos θ with sin θ = x.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q21 1 Mark

Assertion (A): Inverse trigonometric functions are defined only on restricted domains.

Reason (R): Trig functions are not one-one on R, so their inverses require restricted principal-value branches.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Statement-Based Questions 5 questions

Q22 1 Mark

Statement 1: The principal value of cos⁻¹ x lies in [0, π].

Statement 2: The principal value of sin⁻¹ x lies in [−π/2, π/2].

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Correct answer: Option 1 — Both statements are true.

Q23 1 Mark

Statement 1: sin⁻¹(−x) = −sin⁻¹(x) for |x| ≤ 1.

Statement 2: cos⁻¹(−x) = π − cos⁻¹(x) for |x| ≤ 1.

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Correct answer: Option 1 — Both statements are true.

Q24 1 Mark

Statement 1: sin⁻¹(1/2) = π/6.

Statement 2: cos⁻¹(1/2) = π/3.

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Correct answer: Option 1 — Both statements are true.

Q25 1 Mark

Statement 1: 2 tan⁻¹(x) = sin⁻¹(2x/(1+x²)) for |x| ≤ 1.

Statement 2: 2 tan⁻¹(x) = cos⁻¹((1−x²)/(1+x²)) for x ≥ 0.

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Correct answer: Option 1 — Both statements are true.

Q26 1 Mark

Statement 1: sec⁻¹(x) is defined only for x in (−1 1).

Statement 2: sec⁻¹(x) = cos⁻¹(1/x) for |x| ≥ 1.

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Correct answer: Option 4 — Both statements are false.

Case Study / Passage Questions 3 questions

Q27 3 Marks

From a point on the ground 30 m away from the foot of a tower the top of the tower is observed at an angle of elevation of 60°. The observer's eye level coincides with the ground.

The height of the tower equals:

A30 m

B30√3 m

C60 m

D15√3 m
If the angle of elevation is 60° then in radians it equals:

Aπ/6

Bπ/4

Cπ/3

Dπ/2
Express the relationship between the angle of elevation θ, the height h and the distance 30 using inverse tan.

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1. Option 2 — 30√3 m

2. Option 3 — π/3

3. Using tan θ = opposite/adjacent we get tan(π/3) = h/30 so h = 30 tan 60° = 30√3 m. Inverse tangent gives θ = arctan(h/30); knowing h reverses the calculation.

Q28 3 Marks

A camera placed 4 m from a wall is to capture a poster of height 3 m mounted with its base at camera level. Let θ be the angle subtended by the poster at the camera.

The angle θ subtended by the poster at the camera is:

Aarctan(3/4)

Barctan(4/3)

Carcsin(3/5)

Darccos(4/5)
The approximate value of θ is:

A30°

B37° (approx)

C45°

D60°
Why is the inverse tangent used here instead of inverse sine or cosine?

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1. Option 1 — arctan(3/4)

2. Option 2 — 37° (approx)

3. arctan(3/4) ≈ 36.87°. Since 4 m is the horizontal distance and 3 m the vertical extent of the poster the inverse tan gives the angle subtended.

Q29 3 Marks

Two students simplify the expression sin⁻¹(3/5) + sin⁻¹(4/5). One claims the answer is π/2 the other claims π. Verify which student is correct using the identity sin⁻¹ x + sin⁻¹ y = sin⁻¹(x√(1−y²) + y√(1−x²)) for the appropriate range.

The correct value of sin⁻¹(3/5) + sin⁻¹(4/5) is:

Aπ/4

Bπ/3

Cπ/2

Dπ
Using the identity the simplification reduces to:

Asin⁻¹(1) = π/2

Bsin⁻¹(7/5)

Csin⁻¹(0)

Dsin⁻¹(1/2)
Show the calculation step by step using the identity.

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1. Option 3 — π/2

2. Option 1 — sin⁻¹(1) = π/2

3. sin⁻¹(3/5) + sin⁻¹(4/5) = sin⁻¹((3/5)·(3/5) + (4/5)·(4/5)) = sin⁻¹(9/25 + 16/25) = sin⁻¹(1) = π/2. So the first student is correct.

Table-Based Questions 3 questions

Q30 3 Marks

Study the principal values of inverse trigonometric functions:

Function	Principal value branch	Range
sin⁻¹ x	[−π/2, π/2]	[−1, 1]
cos⁻¹ x	[0, π]	[−1, 1]
tan⁻¹ x	(−π/2, π/2)	R
cot⁻¹ x	(0, π)	R
sec⁻¹ x	[0, π] − {π/2}	(−∞,−1] ∪ [1,∞)
cosec⁻¹ x	[−π/2, π/2] − {0}	(−∞,−1] ∪ [1,∞)

The principal value branch of sin⁻¹ x is:

A[0, π]

B[−π/2, π/2]

C(−π/2, π/2)

D(0, π)
The principal value of cos⁻¹(−1/2) is:

Aπ/3

B2π/3

C−π/3

Dπ/6
Why are principal value branches chosen for inverse trig functions?

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1. Option 2 — [−π/2, π/2]

2. Option 2 — 2π/3

3. Inverse trigonometric functions are multi-valued; principal value branches restrict the codomain so that each input has a unique output. The branches are chosen to cover the full range of the original function while keeping the inverse one-one.

Q31 3 Marks

Study the values of inverse trig functions and answer:

x	sin⁻¹ x	cos⁻¹ x	tan⁻¹ x
1	π/2	0	π/4
0	0	π/2	0
−1	−π/2	π	−π/4
1/2	π/6	π/3	—
−1/2	−π/6	2π/3	—

sin⁻¹(1) + cos⁻¹(1) equals:

Aπ/2

Bπ/3

Cπ/4

Dπ
sin⁻¹ x + cos⁻¹ x for x ∈ [−1,1] always equals:

Aπ/2

Bπ

C0

Dπ/3
Verify the identity sin⁻¹ x + cos⁻¹ x = π/2 using x = 1/2.

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1. Option 1 — π/2

2. Option 1 — π/2

3. For all x ∈ [−1,1] the identity sin⁻¹ x + cos⁻¹ x = π/2 holds because sine and cosine are co-functions of complementary angles. The same logic gives tan⁻¹ x + cot⁻¹ x = π/2 for x ∈ R.

Q32 6 Marks

Compute the principal values of the following inverse trigonometric expressions and state the principal value branch used for each.

Expression	Principal value
sin⁻¹(1/2)	?
cos⁻¹(−1/2)	?
tan⁻¹(1)	?
cosec⁻¹(2)	?
sec⁻¹(−2)	?

Picture-Based Questions 1 question

Q33 3 Marks

Study the graphs of y = sin x and y = sin⁻¹ x and answer:

Inverse Trigonometric Functions figure

The two curves are reflections of each other about:

Ax-axis

By-axis

Cy = x

Dorigin
The principal value branch (range) of sin⁻¹ x is:

A[−1, 1]

B[−π/2, π/2]

C[0, π]

DR
Why is the domain of sin x restricted to [−π/2, π/2] before defining sin⁻¹ x?

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1. Option 3 — y = x

2. Option 2 — [−π/2, π/2]

3. Restricting the domain of sin x to [−π/2, π/2] makes it one-one over that interval, so its inverse exists. The range of sin⁻¹ x is exactly this restricted domain, and the graph of the inverse is the reflection of the original about the line y = x.

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