Important Questions Class 12 Maths Chapter 2 cover inverse trigonometric functions, principal value branches, domains, ranges, identities and simplification-based questions. For CBSE Class 12 Mathematics 2026-27, Chapter 2 helps students revise one-mark objective questions, short answers, long answers and case-based applications from Inverse Trigonometric Functions.
Inverse Trigonometric Functions begins with a key idea from functions: an inverse exists only when a function is one-one and onto. Since trigonometric functions are periodic, their natural domains have to be restricted before inverse functions such as sin⁻¹x, cos⁻¹x, tan⁻¹x, cot⁻¹x, sec⁻¹x and cosec⁻¹x can be defined.
This chapter is scoring because many questions depend on principal value branches, domain-range memory and standard identities. These Important Questions Class 12 Maths Chapter 2 include MCQs, very short answers, short answers, long answers, case-based questions and simplification practice for board-style revision.
Key Takeaways
- Principal value branches: Every inverse trigonometric function is taken in its principal branch unless stated otherwise.
- Domain and range: Questions often test whether the given value lies in the correct domain and range.
- Common mistake: sin⁻¹x is not the same as (sin x)⁻¹.
- Board focus: Simplification, principal value, identities and case-based graph/domain questions are important.
Important Questions Class 12 Maths Chapter 2 Exam Pattern Overview
The CBSE Class 12 Mathematics exam is a 3-hour offline paper worth 100 marks, divided into an 80-mark theory paper and 20 marks for internal assessment. The written paper has 38 compulsory questions across objective, short answer, long answer and case-based formats. For Inverse Trigonometric Functions Class 12, students should practise principal value questions, domain-range tables, identity-based simplification and graph-based interpretation.
CBSE Class 12 Maths Theory Paper Pattern
| Section |
Question Type |
No. of Questions |
Marks per Question |
Total Marks |
| Section A |
MCQs and Assertion-Reasoning |
20 |
1 |
20 |
| Section B |
Very Short Answer |
5 |
2 |
10 |
| Section C |
Short Answer |
6 |
3 |
18 |
| Section D |
Long Answer |
4 |
5 |
20 |
| Section E |
Case / Source-Based Questions |
3 |
4 |
12 |
| Total |
|
38 |
|
80 |
Domain and Range of Inverse Trigonometric Functions
This table is the base of Chapter 2. Students should revise it before attempting principal value and simplification questions.
| Function |
Domain |
Principal Value Range |
| sin⁻¹x |
[−1, 1] |
[−π/2, π/2] |
| cos⁻¹x |
[−1, 1] |
[0, π] |
| tan⁻¹x |
R |
(−π/2, π/2) |
| cot⁻¹x |
R |
(0, π) |
| sec⁻¹x |
R − (−1, 1) |
[0, π] − {π/2} |
| cosec⁻¹x |
R − (−1, 1) |
[−π/2, π/2] − {0} |
MCQs and Objective Questions from Class 12 Maths Chapter 2
These questions test principal values, branch ranges and basic conceptual accuracy. Write the answer only after checking the principal value branch.
Q1. The principal value of sin⁻¹(−1/2) is:
(a) π/6
(b) −π/6
(c) 5π/6
(d) −5π/6
Answer: (b) −π/6
Since sin(−π/6) = −1/2 and −π/6 lies in [−π/2, π/2], the principal value is −π/6.
Q2. The range of cos⁻¹x is:
(a) [−π/2, π/2]
(b) (−π/2, π/2)
(c) [0, π]
(d) (0, π)
Answer: (c) [0, π]
The principal value branch of cos⁻¹x has range [0, π].
Q3. sin⁻¹x should not be confused with:
(a) cosec x
(b) (sin x)⁻¹
(c) sin x
(d) cos⁻¹x
Answer: (b) (sin x)⁻¹
sin⁻¹x means inverse sine or arc sine. It does not mean reciprocal of sin x.
Q4. tan⁻¹x is defined for:
(a) [−1, 1]
(b) R
(c) R − (−1, 1)
(d) [0, π]
Answer: (b) R
The domain of tan⁻¹x is all real numbers.
Q5. The principal value of cot⁻¹(−1/√3) is:
(a) π/3
(b) 2π/3
(c) −π/3
(d) 5π/6
Answer: (b) 2π/3
The range of cot⁻¹x is (0, π). Since cot(2π/3) = −1/√3, the answer is 2π/3.
Q6. The principal value branch of sec⁻¹x is:
(a) [−π/2, π/2]
(b) [0, π] − {π/2}
(c) (−π/2, π/2)
(d) (0, π)
Answer: (b) [0, π] − {π/2}
The function sec x is not defined at π/2 in this branch.
Very Short Answer Questions on Principal Values
These 2-mark questions usually need one substitution and one branch check.
Q7. Find the principal value of cos⁻¹(√3/2).
Answer: π/6
Since cos(π/6) = √3/2 and π/6 lies in [0, π], the principal value is π/6.
Q8. Find the principal value of tan⁻¹(−1).
Answer: −π/4
Since tan(−π/4) = −1 and −π/4 lies in (−π/2, π/2), the principal value is −π/4.
Q9. Find the principal value of cosec⁻¹(2).
Answer: π/6
Let cosec⁻¹(2) = y.
Then cosec y = 2, so sin y = 1/2.
The principal value is π/6.
Q10. Find the value of cos⁻¹(−1/2).
Answer: 2π/3
Since cos(2π/3) = −1/2 and 2π/3 lies in [0, π], the principal value is 2π/3.
Q11. Find the value of cot⁻¹(√3).
Answer: π/6
Since cot(π/6) = √3 and π/6 lies in (0, π), the principal value is π/6.
Q12. Find the value of sec⁻¹(2).
Answer: π/3
Since sec y = 2 means cos y = 1/2.
In the principal range [0, π] − {π/2}, y = π/3.
Short Answer Questions on Inverse Trigonometric Functions Class 12
These 3-mark questions need simplification, identity use or correct interpretation of principal branches.
Q13. Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
Answer:
tan⁻¹(1) = π/4
cos⁻¹(−1/2) = 2π/3
sin⁻¹(−1/2) = −π/6
So,
= π/4 + 2π/3 − π/6
= π/4 + π/2
= 3π/4
Therefore, the value is 3π/4.
Q14. Find the value of cos⁻¹(1/2) + 2 sin⁻¹(1/2).
Answer:
cos⁻¹(1/2) = π/3
sin⁻¹(1/2) = π/6
So,
cos⁻¹(1/2) + 2 sin⁻¹(1/2)
= π/3 + 2(π/6)
= π/3 + π/3
= 2π/3
Therefore, the value is 2π/3.
Q15. If sin⁻¹x = y, write the range of y.
Answer:
If sin⁻¹x = y, then y lies in the principal value range of sin⁻¹x.
So,
−π/2 ≤ y ≤ π/2
Also, x must lie in [−1, 1].
Q16. Find the value of sin(sin⁻¹(3/5)).
Answer:
For x ∈ [−1, 1],
sin(sin⁻¹x) = x
Therefore,
sin(sin⁻¹(3/5)) = 3/5
Q17. Find the value of sin⁻¹(sin(3π/5)).
Answer:
The principal range of sin⁻¹x is [−π/2, π/2].
3π/5 does not lie in this range.
sin(3π/5) = sin(π − 3π/5) = sin(2π/5)
Since 2π/5 lies in [−π/2, π/2],
sin⁻¹(sin(3π/5)) = 2π/5
Q18. Write tan⁻¹(tan(7π/6)) in principal value form.
Answer:
The principal range of tan⁻¹x is (−π/2, π/2).
tan(7π/6) = tan(π + π/6) = tan(π/6)
So,
tan⁻¹(tan(7π/6)) = tan⁻¹(tan(π/6)) = π/6
Long Answer Questions on Identities and Simplification
These questions often appear as proof or simplification problems. Students should mention substitutions and principal value conditions where required.
Q19. Prove that sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹x, where −1/2 ≤ x ≤ 1/2.
Answer:
Let x = sin θ.
Then,
sin⁻¹x = θ
Since −1/2 ≤ x ≤ 1/2, θ lies in [−π/6, π/6].
So 2θ lies in [−π/3, π/3], which is within the principal range of sin⁻¹x.
Now,
2x√(1 − x²)
= 2 sin θ √(1 − sin²θ)
= 2 sin θ cos θ
= sin 2θ
Therefore,
sin⁻¹(2x√(1 − x²))
= sin⁻¹(sin 2θ)
= 2θ
= 2 sin⁻¹x
Hence proved.
Q20. Prove that cos⁻¹(4x³ − 3x) = 3 cos⁻¹x, where 1/2 ≤ x ≤ 1.
Answer:
Let x = cos θ.
Then,
cos⁻¹x = θ
Since 1/2 ≤ x ≤ 1, θ lies in [0, π/3].
So 3θ lies in [0, π], which is the principal range of cos⁻¹x.
Now,
4x³ − 3x
= 4 cos³θ − 3 cos θ
= cos 3θ
Therefore,
cos⁻¹(4x³ − 3x)
= cos⁻¹(cos 3θ)
= 3θ
= 3 cos⁻¹x
Hence proved.
Q21. Write tan⁻¹((1 − cos x)/sin x) in simplest form, where 0 < x < π.
Answer:
We know:
1 − cos x = 2 sin²(x/2)
sin x = 2 sin(x/2) cos(x/2)
So,
(1 − cos x)/sin x
= 2 sin²(x/2) / 2 sin(x/2) cos(x/2)
= tan(x/2)
Therefore,
tan⁻¹((1 − cos x)/sin x)
= tan⁻¹(tan(x/2))
Since 0 < x < π, we get 0 < x/2 < π/2.
So,
tan⁻¹(tan(x/2)) = x/2
Hence, the simplest form is x/2.
Q22. Write cot⁻¹(1/√(x² − 1)) in simplest form, where x > 1.
Answer:
Let x = sec θ.
Then,
√(x² − 1) = √(sec²θ − 1) = tan θ
So,
1/√(x² − 1) = 1/tan θ = cot θ
Therefore,
cot⁻¹(1/√(x² − 1))
= cot⁻¹(cot θ)
= θ
Since x = sec θ,
θ = sec⁻¹x
Hence, the simplest form is sec⁻¹x.
Q23. Find the value of tan⁻¹(√3) − cot⁻¹(−√3).
Answer:
tan⁻¹(√3) = π/3
For cot⁻¹(−√3), the range is (0, π).
cot(5π/6) = −√3
So,
cot⁻¹(−√3) = 5π/6
Therefore,
tan⁻¹(√3) − cot⁻¹(−√3)
= π/3 − 5π/6
= 2π/6 − 5π/6
= −3π/6
= −π/2
Therefore, the value is −π/2.
Case-Based Questions from Class 12 Maths Chapter 2
Case-based questions from this chapter usually test graphs, branch restrictions and real-life use of inverse trigonometric functions.
Q24. Case-Based Question: Principal Branch and Graph Reflection
A student draws the graph of y = sin x over the interval [−π/2, π/2]. The function is one-one and onto from [−π/2, π/2] to [−1, 1]. The graph of y = sin⁻¹x is obtained by reflecting this restricted sine graph in the line y = x.
(a) Why is the domain of sin x restricted before defining sin⁻¹x?
The sine function is not one-one over R. It must be restricted to a domain where it becomes one-one and onto.
(b) What is the principal value range of sin⁻¹x?
The principal value range is [−π/2, π/2].
(c) What is the domain of sin⁻¹x?
The domain is [−1, 1].
(d) How is the graph of y = sin⁻¹x obtained from y = sin x?
It is obtained by reflecting the graph of y = sin x in the line y = x, after restricting sine to its principal branch.
Q25. Case-Based Question: Height and Angle Measurement
An engineer uses an angle-measuring device to calculate the angle of elevation of the top of a tower. If the ratio of height to horizontal distance is x, then the angle can be written as tan⁻¹x. The principal value branch of tan⁻¹x is used to get a unique angle.
(a) What is the domain of tan⁻¹x?
The domain is R.
(b) What is the principal value range of tan⁻¹x?
The range is (−π/2, π/2).
(c) If height/distance = 1, what is the angle?
tan⁻¹(1) = π/4.
(d) Why does tan⁻¹x give a unique angle in this case?
Because tan x is restricted to (−π/2, π/2), where it is one-one and onto.
Q26. Case-Based Question: Reciprocal Trigonometric Functions
A student writes sec⁻¹x and cosec⁻¹x while solving a problem. The teacher asks the student to check the domain before substituting values.
(a) What is the domain of sec⁻¹x?
The domain is R − (−1, 1).
(b) What is the principal value range of sec⁻¹x?
The range is [0, π] − {π/2}.
(c) What is the domain of cosec⁻¹x?
The domain is R − (−1, 1).
(d) Why is x = 1/2 not allowed for sec⁻¹x?
Because 1/2 lies in (−1, 1), which is excluded from the domain of sec⁻¹x.
Practice Questions on Domain, Range and Principal Branch
This group is useful for quick revision before attempting mixed exercise problems.
Q27. State the domain and range of cot⁻¹x.
Answer:
Domain of cot⁻¹x = R
Principal value range of cot⁻¹x = (0, π)
Q28. State the domain and range of cosec⁻¹x.
Answer:
Domain of cosec⁻¹x = R − (−1, 1)
Principal value range = [−π/2, π/2] − {0}
Q29. Find the principal value of sec⁻¹(−2).
Answer:
Let sec⁻¹(−2) = y.
Then,
sec y = −2
So, cos y = −1/2
In the principal range [0, π] − {π/2}, cos y = −1/2 at y = 2π/3.
Therefore,
sec⁻¹(−2) = 2π/3
Q30. Find the value of sin(tan⁻¹x), where |x| < 1.
Answer:
Let tan⁻¹x = θ.
Then,
tan θ = x = opposite/adjacent = x/1
So, hypotenuse = √(1 + x²)
Therefore,
sin θ = opposite/hypotenuse = x/√(1 + x²)
Hence,
sin(tan⁻¹x) = x/√(1 + x²)
Important Questions Class 12 Maths Chapter-Wise