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Chapter 2: Inverse Trig Functions

NCERT Solutions for CBSE Class 10 Mathematics — 7 solved questions with detailed explanations.

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Questions

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Solved Questions

Q1. The principal value of sin⁻¹(1/2) is:

π/6
π/3
π/4
5π/6

Difficulty: Easy · Topic: Principal Values

Solution

sin(π/6) = 1/2. Principal value of sin⁻¹ lies in [-π/2, π/2], so answer is π/6.

Q2. sin⁻¹(x) + cos⁻¹(x) =:

π/2
1
0
π

Difficulty: Easy-Medium · Topic: Inverse Trig Functions

Solution

π/2.

Q3. tan⁻¹(x) + tan⁻¹(1/x) =:

1
0
π/2 (x>0)
π

Difficulty: Easy-Medium · Topic: Inverse Trig Functions

Solution

π/2 (x>0).

Q4. The range of sin⁻¹(x) is:

[-π/2, π/2]
(-π, π)
[0, 2π]
[0, π]

Difficulty: Easy-Medium · Topic: Inverse Trig Functions

Solution

[-π/2, π/2].

Q5. The range of cos⁻¹(x) is:

[-π/2, π/2]
[0, 2π]
(-π, π)
[0, π]

Difficulty: Easy-Medium · Topic: Inverse Trig Functions

Solution

[0, π].

Q6. The range of tan⁻¹(x) is:

[-π/2, π/2]
(-π/2, π/2)
(-π, π)
[0, π]

Difficulty: Easy-Medium · Topic: Inverse Trig Functions

Solution

(-π/2, π/2).

Q7. tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) equals:

π
3π/4
π/4
2π

Difficulty: Medium · Topic: Inverse Trig Sums

Solution

tan⁻¹(1) = π/4. Using tan⁻¹(2) + tan⁻¹(3) = π + tan⁻¹((2+3)/(1-6)) = π + tan⁻¹(-1) = π - π/4 = 3π/4. Total = π/4 + 3π/4 = π.

Other Chapters in Mathematics

Ch 1: Relations & Functions Ch 3: Matrices Ch 4: Determinants Ch 5: Continuity & Differentiability Ch 6: Applications of Derivatives Ch 7: Integrals Ch 8: Applications of Integrals Ch 9: Differential Equations Ch 10: Vector Algebra Ch 11: Three Dimensional Geometry Ch 12: Linear Programming Ch 13: Probability

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