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

Working through inverse trigonometric functions means getting comfortable with principal value branches and domains, and this set of questions puts that understanding to the test.

Last Updated: September 23, 2026

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.

Frequently Asked Questions

How would you find the value of sin inverse of one half?
Since sin(30 degrees) = one half, sin inverse of one half = pi over 6 or 30 degrees, within the principal range.

How would you simplify tan inverse(1) + tan inverse(2) + tan inverse(3)?
Using known values and the addition formula, this expression simplifies to pi.

Written by Satish

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