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.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Inverse Trigonometric Functions" — 8 important questions with detailed answers for…
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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)).
- + 5 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 |
Showing 3 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 principa...
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:
t...
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 th...
Showing 3 of 8 questions. Visit the full page for complete solutions.