Inverse Trigonometric Functions — Class 12 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Inverse Trigonometric Functions" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Inverse Trigonometric Functions" — 8 important questions with detailed answers for…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Find the principal value of sin⁻¹(-1/2).
Prove that tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0.
Find the value of cos⁻¹(cos(7π/6)).
Find the principal value of tan⁻¹(√3).
Simplify: sin⁻¹(3/5) + sin⁻¹(4/5).
If sin⁻¹(x) = π/6, find the value of x.
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Find the principal value of sin⁻¹(-1/2).	✓ Solved
Prove that tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0.	✓ Solved
Find the value of cos⁻¹(cos(7π/6)).	✓ Solved
Find the principal value of tan⁻¹(√3).	✓ Solved
Simplify: sin⁻¹(3/5) + sin⁻¹(4/5).	✓ Solved
If sin⁻¹(x) = π/6, find the value of x.	✓ Solved

Showing 6 of 8 questions

Q1: Find the principal value of sin⁻¹(-1/2).

We need to find the principal value of sin⁻¹(-1/2). Let θ = sin⁻¹(-1/2) Then sin(θ) = -1/2 The range of sin⁻¹ is [-π/2, π/2]. We know sin(π/6) = 1/2. Since we need a negative value and the principal range is [-π/2, π/2]: sin(-π/6) = -sin(π/6) = -1/2 Since -π/6 ∈ [-π/2, π/2], this is the principal value. CONCLUSION: sin⁻¹(-1/2) = -π/6

Q2: Prove that tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0.

We need to prove: tan⁻¹(x) + tan⁻¹(1/x) = π/2 for x > 0 Let α = tan⁻¹(x) and β = tan⁻¹(1/x), where x > 0. Then tan(α) = x and tan(β) = 1/x Also, 0 < α < π/2 and 0 < β < π/2 (since x > 0) We compute tan(α + β): tan(α + β) = (tan(α) + tan(β))/(1 - tan(α)tan(β)) Substituting: tan(α + β) = (x + 1/x)/(1 - x(1/x)) = (x + 1/x)/(1 - 1) = (x + 1/x)/0 = undefined tan(α + β) is undefined, which means α + β = π/2 (since α + β ∈ (0, π)). Therefore, ...

Q3: Find the value of cos⁻¹(cos(7π/6)).

We need to find cos⁻¹(cos(7π/6)). The range of cos⁻¹ is [0, π]. First, let's find cos(7π/6): 7π/6 = π + π/6 cos(7π/6) = cos(π + π/6) = -cos(π/6) = -√3/2 Now we need to find cos⁻¹(-√3/2). This means finding θ ∈ [0, π] such that cos(θ) = -√3/2. We know cos(π/6) = √3/2. For the negative value in the range [0, π]: cos(π - π/6) = cos(5π/6) = -cos(π/6) = -√3/2 Since 5π/6 ∈ [0, π], we have: cos⁻¹(-√3/2) = 5π/6 CONCLUSION: cos⁻¹(cos(7π/6)) = 5π/6

Q4: Find the principal value of tan⁻¹(√3).

We need to find the principal value of tan⁻¹(√3). Let θ = tan⁻¹(√3) Then tan(θ) = √3 The range of tan⁻¹ is (-π/2, π/2). We know that tan(π/3) = √3. Since π/3 ∈ (-π/2, π/2), we have: CONCLUSION: tan⁻¹(√3) = π/3

Q5: Simplify: sin⁻¹(3/5) + sin⁻¹(4/5).

Let α = sin⁻¹(3/5) and β = sin⁻¹(4/5) Then sin(α) = 3/5 and sin(β) = 4/5, where α, β ∈ [0, π/2] (since both values are positive) Finding cos(α) and cos(β): cos(α) = √(1 - sin²(α)) = √(1 - 9/25) = √(16/25) = 4/5 (positive in first quadrant) cos(β) = √(1 - sin²(β)) = √(1 - 16/25) = √(9/25) = 3/5 (positive in first quadrant) Finding sin(α + β): sin(α + β) = sin(α)cos(β) + cos(α)sin(β) = (3/5)(3/5) + (4/5)(4/5) = 9/25 + 16/25 = 25/25 = 1 Since sin(α + ...

Q6: If sin⁻¹(x) = π/6, find the value of x.

Given: sin⁻¹(x) = π/6 By definition of inverse sine function: if sin⁻¹(x) = π/6, then sin(π/6) = x We know that sin(π/6) = 1/2 Therefore: x = 1/2 VERIFICATION: sin⁻¹(1/2) = π/6 ✓ (since sin(π/6) = 1/2 and π/6 is in the range [-π/2, π/2]) CONCLUSION: x = 1/2

Showing 6 of 8 questions. Visit the full page for complete solutions.