Inverse trigonometric functions answer the reverse question: given a ratio, which angle produced it? Chapter 2 works through the NCERT exercise problems on evaluating these functions, their principal value branches, and key identities.
Last Updated: September 23, 2026
NCERT Exercise 2.1 Solutions (All 14 Questions)
Note: under the 2023 CBSE syllabus rationalisation, Exercise 2.2 and the Miscellaneous Exercise of this chapter were removed; Exercise 2.1 (14 questions, principal values) is the only exercise remaining in the current 2026-27 syllabus.
1. Find the principal value of sin⁻¹(−1/2).
Ans: sin⁻¹(−1/2) = −π/6, since sin(−π/6) = −1/2 and −π/6 lies in the principal value range [−π/2, π/2].
2. Find the principal value of cos⁻¹(√3/2).
Ans: cos⁻¹(√3/2) = π/6, since cos(π/6) = √3/2 and π/6 lies in [0, π].
3. Find the principal value of cosec⁻¹(2).
Ans: cosec⁻¹(2) = π/6, since cosec(π/6) = 2.
4. Find the principal value of tan⁻¹(−√3).
Ans: tan⁻¹(−√3) = −π/3, since tan(−π/3) = −√3.
5. Find the principal value of cos⁻¹(−1/2).
Ans: cos⁻¹(−1/2) = 2π/3, since cos(2π/3) = −1/2 and 2π/3 ∈ [0, π].
6. Find the principal value of tan⁻¹(−1).
Ans: tan⁻¹(−1) = −π/4.
7. Find the principal value of sec⁻¹(2/√3).
Ans: sec⁻¹(2/√3) = π/6, since sec(π/6) = 2/√3.
8. Find the principal value of cot⁻¹(√3).
Ans: cot⁻¹(√3) = π/6, since cot(π/6) = √3.
9. Find the principal value of cos⁻¹(−1/√2).
Ans: cos⁻¹(−1/√2) = 3π/4, since cos(3π/4) = −1/√2 and 3π/4 ∈ [0, π].
10. Find the principal value of cosec⁻¹(−√2).
Ans: cosec⁻¹(−√2) = −π/4, since cosec(−π/4) = −√2 and −π/4 lies in the principal value range [−π/2, π/2]−{0}.
11. Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
Ans: π/4 + 2π/3 + (−π/6) = 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4.
12. Find the value of cos⁻¹(1/2) + 2sin⁻¹(1/2).
Ans: π/3 + 2(π/6) = π/3 + π/3 = 2π/3.
13. If sin⁻¹x = y, then: (A) 0≤y≤π (B) −π/2≤y≤π/2 (C) 0<y<π (D) −π/2<y<π/2.
Ans: (B) −π/2≤y≤π/2 — this is exactly the principal value branch of sin⁻¹x, which is a closed interval (endpoints included).
14. tan⁻¹(√3) − sec⁻¹(−2) is equal to: (A) π (B) −π/3 (C) π/3 (D) 2π/3.
Ans: (B) −π/3, since tan⁻¹(√3)=π/3 and sec⁻¹(−2)=π−sec⁻¹(2)=π−π/3=2π/3, giving π/3−2π/3=−π/3.
Class 12 Mathematics Chapter 2 – Notes and Extra Questions
Along with these NCERT Solutions, students can also use the Class 12 Mathematics Chapter 2 Extra Questions and Class 12 Mathematics Chapter 2 Revision Notes for quick revision and extra practice.
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