Inverse Trigonometric Functions — Class 12 MCQs with Answers

Inverse Trigonometric Functions — important questions & MCQs with answers (Class 12 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Range of sin⁻¹x:

   A[0, π]B[−π/2, π/2]CRD[0, 2π]
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   Answer: [−π/2, π/2]

   Explanation: The principal-value branch restricts sin⁻¹x's output to [−π/2, π/2], not all of R — students confuse this with the domain [−1, 1], which is the input restriction, not the output.

2. Q2 of 6Easy

   Range of sin⁻¹ x:

   A[0, π]B[−π/2, π/2]C[−π, π]DR
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   Answer: [−π/2, π/2]

   Explanation: sin⁻¹x's principal branch outputs only [−π/2, π/2] — half of cos⁻¹x's [0, π] range — because sine is one-to-one only on that restricted domain.

3. Q3 of 6Medium

   The principal value of sin⁻¹(−1/2) is:

   A−5π/6Bπ/6C7π/6D−π/6
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   Answer: −π/6

   Explanation: Within [−π/2, π/2] the angle whose sine is −1/2 is −π/6.

4. Q4 of 6Medium

   Assertion (A): tan⁻¹(√3) − sec⁻¹(−2) = −π/3. Reason (R): The principal value of tan⁻¹(√3) is π/3 and the principal value of sec⁻¹(−2) is 2π/3.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: R is true: tan(π/3) = √3 with π/3 in (−π/2, π/2), and sec(2π/3) = −2 with 2π/3 in [0, π] − {π/2}. Using these two values, π/3 − 2π/3 = −π/3, so A is true and follows directly from R.

5. Q5 of 6Hard

   tan⁻¹(1/2) + tan⁻¹(1/3) =

   Aπ/2Bπ/4Cπ/6Dπ/3
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   Answer: π/4

   Explanation: Product xy = (1/2)(1/3) = 1/6 < 1, so the direct formula applies: tan(A+B) = (1/2+1/3)/(1−1/6) = (5/6)/(5/6) = 1, giving A+B = π/4.

6. Q6 of 6Hard

   Assertion (A): There is no real number x for which sin⁻¹x = 2π/3. Reason (R): The range of the principal value branch of sin⁻¹ is [−π/2, π/2].

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: R is true. Since 2π/3 > π/2, the value 2π/3 is not in the range of sin⁻¹, so no x gives it (even though sin(2π/3) = √3/2, sin⁻¹(√3/2) is π/3). A is true, and R is exactly the reason.

Inverse Trigonometric Functions notes: key concepts (Class 12 Mathematics)

Trigonometric functions are made invertible by restricting each to a principal-value branch: sin⁻¹ and cos⁻¹ take domain [−1, 1], tan⁻¹ takes all of R. The chapter fixes these domains and ranges, drills standard principal values, and applies complementary, reciprocal, sum and double-angle identities to evaluate expressions.

Principal value branch

The restricted interval on which a trig function is one-one, so an inverse exists; sin⁻¹ uses [−π/2, π/2], cos⁻¹ uses [0, π].

Domain of sin⁻¹, cos⁻¹

Both are defined only for x ∈ [−1, 1], since sine and cosine never exceed 1 in magnitude.

Range of sin⁻¹

sin⁻¹x returns an angle in [−π/2, π/2]; this is why sin⁻¹(1/2) = π/6 and never 5π/6.

Range of cos⁻¹

cos⁻¹x returns an angle in [0, π]; it is never negative, so cos⁻¹(−1/2) = 2π/3.

Range of tan⁻¹

tan⁻¹x is defined for all real x and outputs an angle in the open interval (−π/2, π/2).

sec⁻¹ and csc⁻¹ domains

Both require |x| ≥ 1; sec⁻¹ ranges over [0, π]\{π/2} and csc⁻¹ over [−π/2, π/2]\{0}.

Complementary identities

sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2, sec⁻¹x + csc⁻¹x = π/2 across their common domains.

Reciprocal relations

cot⁻¹x = tan⁻¹(1/x) for x > 0; sec⁻¹x = cos⁻¹(1/x); csc⁻¹x = sin⁻¹(1/x) for |x| ≥ 1.

Sum formula for tan⁻¹

tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) holds only when xy < 1; otherwise adjust by ±π.

Double-angle forms

2 tan⁻¹x = tan⁻¹(2x/(1−x²)) = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²)), each on its valid range.

Self-composition rules

sin(sin⁻¹x) = x for x ∈ [−1, 1], but sin⁻¹(sin x) = x only when x lies in the principal branch.

Negative arguments

sin⁻¹(−x) = −sin⁻¹x and tan⁻¹(−x) = −tan⁻¹x, but cos⁻¹(−x) = π − cos⁻¹x — the sign rule differs.