Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions – Extra Questions with Answers

Extra practice questions for Class 12 Maths Chapter 2 (Inverse Trigonometric Functions), beyond the textbook. These Class 12 Mathematics Chapter 2 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. What is the range of cos⁻¹x?
Ans: [0, π].

Q2. Find the principal value of sin⁻¹(1/2).
Ans: π/6.

Q3. Find the principal value of tan⁻¹(1).
Ans: π/4.

Q4. Is sin⁻¹x an odd or even function?
Ans: Odd function (sin⁻¹(−x)=−sin⁻¹x).

Q5. What is the domain of sec⁻¹x?
Ans: R−(−1,1), i.e. |x|≥1.

Short Answer Questions (2–3 marks)

Q6. Find the value of cos⁻¹(−1/2).
Ans: cos⁻¹(−1/2)=π−cos⁻¹(1/2)=π−π/3=2π/3.

Q7. Prove that tan⁻¹(1)+tan⁻¹(2)+tan⁻¹(3)=π.
Ans: Using tan⁻¹x+tan⁻¹y=π+tan⁻¹((x+y)/(1−xy)) when xy>1: tan⁻¹(2)+tan⁻¹(3)=π+tan⁻¹((2+3)/(1−6))=π+tan⁻¹(−1)=π−π/4=3π/4. Adding tan⁻¹(1)=π/4: total=3π/4+π/4=π.

Q8. Simplify sin(cos⁻¹x) in terms of x.
Ans: If cos⁻¹x=θ, then cosθ=x, so sinθ=√(1−x²) (since θ∈[0,π], sinθ≥0). So sin(cos⁻¹x)=√(1−x²).

Higher-Order Thinking / Application Questions

Q9. Prove that 2tan⁻¹(1/2)+tan⁻¹(1/7)=tan⁻¹(31/17), showing all steps, and explain why care must be taken with the sum formula’s domain restriction throughout.
Ans: First, find 2tan⁻¹(1/2) using the double angle-like formula tan⁻¹x+tan⁻¹x=tan⁻¹(2x/(1−x²)) valid since x²=1/4<1: 2tan⁻¹(1/2)=tan⁻¹((2×1/2)/(1−1/4))=tan⁻¹(1/(3/4))=tan⁻¹(4/3). Now add tan⁻¹(4/3)+tan⁻¹(1/7): since xy=(4/3)(1/7)=4/21<1, we use the direct sum formula: tan⁻¹((4/3+1/7)/(1−4/21))=tan⁻¹((28/21+3/21)/(17/21))=tan⁻¹((31/21)/(17/21))=tan⁻¹(31/17). This confirms the identity. Throughout, care must be taken because the addition formula tan⁻¹x+tan⁻¹y=tan⁻¹((x+y)/(1−xy)) is only valid (giving a result directly, without needing to add or subtract π) when xy<1; if xy>1 an extra ±π term must be added depending on the signs of x and y, since without this adjustment the formula could give a value outside the correct principal value range of tan⁻¹.

Q10. Explain, using the concept of function invertibility, why we cannot simply say sin⁻¹(sin x)=x for all real x, and specify the exact condition under which this identity does hold.
Ans: The function sin(x) is periodic (period 2π) and is not one-to-one over its entire domain (all real numbers) — multiple different x values (e.g. x=0, x=π, x=2π, etc.) can produce the same sin(x) output, meaning sin(x) has no true inverse function over all of R. To define an inverse, mathematicians restrict sin(x) to a specific interval where it IS one-to-one (strictly increasing/decreasing and covers the full range [−1,1] exactly once), which is chosen to be [−π/2, π/2] — this restricted version is what sin⁻¹x is actually the inverse of. Consequently, the identity sin⁻¹(sin x)=x only holds true when x itself already lies within this specific principal value domain, i.e. when x∈[−π/2, π/2]. If x lies outside this interval (e.g. x=3π/4), then sin⁻¹(sin x) will NOT simply return x, but instead will return some other value within [−π/2, π/2] that has the same sine value as x (for x=3π/4, sin(3π/4)=√2/2, and sin⁻¹(√2/2)=π/4, not 3π/4) — this is a direct consequence of sin⁻¹x being defined as the inverse of the domain-restricted sine function, not the full periodic sine function, so the identity sin⁻¹(sin x)=x is only guaranteed within the principal value domain of sin(x) itself.

Written by Satish

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