RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
Trigonometric functions are periodic, so they are not invertible on their whole domain — inverse trig functions exist only once we restrict to a principal value branch. This chapter fixes those branches, links the six inverse functions through complementary and reciprocal identities, and builds addition formulas that let you simplify expressions like tan⁻¹x + tan⁻¹y without ever computing an angle numerically.
The value of sin⁻¹(sin(3π/4)) is:
Answer
The value is π/4. 3π/4 does not lie in the principal value range [−π/2, π/2] of sin⁻¹, so it must be reduced. sin(3π/4) = sin(π − 3π/4) = sin(π/4). Since π/4 does lie in [−π/2, π/2], sin⁻¹(sin(3π/4)) = π/4.
The value of tan⁻¹(1) + cos⁻¹(−1/2) is:
Answer
The value is 11π/12. tan⁻¹(1) = π/4, since tan(π/4) = 1 and π/4 lies in (−π/2, π/2). cos⁻¹(−1/2) = 2π/3, since cos(2π/3) = −1/2 and 2π/3 lies in [0, π]. Adding: π/4 + 2π/3 = 3π/12 + 8π/12 = 11π/12.
The domain of the function cos⁻¹(2x − 1) is:
Answer
The domain is [0,1]. cos⁻¹ is defined only for inputs in [−1,1], so we need −1 ≤ 2x − 1 ≤ 1. Adding 1 throughout: 0 ≤ 2x ≤ 2, and dividing by 2: 0 ≤ x ≤ 1. So the domain is [0,1].
The value of sin(cos⁻¹(3/5)) is:
Answer
The value is 4/5. Let θ = cos⁻¹(3/5), so cosθ = 3/5 with θ in [0,π], meaning sinθ ≥ 0. Using sin²θ + cos²θ = 1: sin²θ = 1 − 9/25 = 16/25, so sinθ = 4/5. Hence sin(cos⁻¹(3/5)) = 4/5.
Assertion (A): The principal value of sec⁻¹(−2) is 2π/3. Reason (R): The range of sec⁻¹ is [0,π] − {π/2}.
Answer
Both A and R are true and R is the correct explanation of A. R correctly states the principal value range of sec⁻¹. For A: sec⁻¹(−2) = θ means secθ = −2, i.e. cosθ = −1/2, with θ restricted to [0,π] − {π/2}. cos(2π/3) = −1/2, and 2π/3 lies in the allowed range, so sec⁻¹(−2) = 2π/3. This confirms A, and it relies directly on the range stated in R.
Find the value of cot⁻¹(−√3), expressing your answer as a principal value.
Answer
Let θ = cot⁻¹(−√3), so cotθ = −√3 with θ restricted to (0, π). Since cotangent is negative, θ must lie in the second quadrant portion of (0,π). cot(5π/6) = cos(5π/6)/sin(5π/6) = (−√3/2)/(1/2) = −√3. Since 5π/6 lies in (0,π), cot⁻¹(−√3) = 5π/6.
Prove that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4.
Answer
Use the identity tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)), valid here since xy = (1/2)(1/3) = 1/6 < 1. Compute x + y = 1/2 + 1/3 = 5/6, and 1 − xy = 1 − 1/6 = 5/6. So tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((5/6)/(5/6)) = tan⁻¹(1). tan⁻¹(1) = π/4, since tan(π/4) = 1 and π/4 lies in (−π/2, π/2). Hence tan⁻¹(1/2) + tan⁻¹(1/3) = π/4, as required.
Simplify: sin⁻¹(sin(7π/6)).
Answer
7π/6 does not lie in the principal range [−π/2, π/2], so it must be reduced using the sine identity for angles beyond π. sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2. Now sin⁻¹(−1/2) = −π/6, since sin(−π/6) = −1/2 and −π/6 lies in [−π/2, π/2]. So sin⁻¹(sin(7π/6)) = −π/6.
Solve for x: tan⁻¹((x−1)/(x−2)) + tan⁻¹((x+1)/(x+2)) = π/4.
Answer
Apply the addition formula tan⁻¹A + tan⁻¹B = tan⁻¹((A+B)/(1−AB)), so we need (A+B)/(1−AB) = tan(π/4) = 1. A + B = (x−1)/(x−2) + (x+1)/(x+2) = [(x−1)(x+2) + (x+1)(x−2)] / [(x−2)(x+2)]. (x−1)(x+2) = x² + x − 2, and (x+1)(x−2) = x² − x − 2. Their sum is 2x² − 4. So A + B = (2x² − 4)/(x² − 4). 1 − AB = 1 − [(x−1)(x+1)]/[(x−2)(x+2)] = 1 − (x²−1)/(x²−4) = [(x²−4) − (x²−1)]/(x²−4) = −3/(x²−4). So (A+B)/(1−AB) = [(2x²−4)/(x²−4)] ÷ [−3/(x²−4)] = (2x²−4)/(−3). Setting this equal to 1: (2x²−4)/(−3) = 1 ⇒ 2x² − 4 = −3 ⇒ 2x² = 1 ⇒ x² = 1/2. So x = ±1/√2. Both values keep x away from 2 and −2, so both are valid solutions.
Prove that: sin⁻¹(3/5) − sin⁻¹(8/17) = cos⁻¹(84/85).
Answer
Let A = sin⁻¹(3/5), so sinA = 3/5. Since A ∈ [−π/2,π/2] and sine is positive, cosA = √(1 − 9/25) = 4/5. Let B = sin⁻¹(8/17), so sinB = 8/17 and cosB = √(1 − 64/289) = 15/17. Using sin(A−B) = sinA cosB − cosA sinB: sin(A−B) = (3/5)(15/17) − (4/5)(8/17) = 45/85 − 32/85 = 13/85. Using cos(A−B) = cosA cosB + sinA sinB: cos(A−B) = (4/5)(15/17) + (3/5)(8/17) = 60/85 + 24/85 = 84/85. Since cos(A−B) = 84/85 and A−B lies in the range where cos⁻¹ is defined and single-valued for this configuration, A − B = cos⁻¹(84/85). Therefore sin⁻¹(3/5) − sin⁻¹(8/17) = cos⁻¹(84/85), as required.
Read the following and answer the questions that follow: A security camera is mounted on top of a pole of height 15 m. A car is on the ground at a horizontal distance x metres from the base of the pole. The angle of elevation θ from the base of the pole to the camera, as seen relative to the line joining the base to the car, can be modelled as θ(x) = tan⁻¹(15/x) for x > 0. (a) Find θ when x = 15 m. (b) Find θ when x = 15√3 m. (c) Write θ as a function of x in general. (d) What happens to θ as x becomes very large (x → ∞)?
Answer
(a) θ(15) = tan⁻¹(15/15) = tan⁻¹(1) = π/4. (b) θ(15√3) = tan⁻¹(15/(15√3)) = tan⁻¹(1/√3) = π/6. (c) In general, θ(x) = tan⁻¹(15/x), giving the angle for any horizontal distance x > 0. (d) As x → ∞, the ratio 15/x → 0, so θ(x) = tan⁻¹(15/x) → tan⁻¹(0) = 0. The angle shrinks towards 0 as the car moves very far away.
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