Inverse Trigonometric Functions Class 12 Notes - CBSE Maths Chapter 2

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Chapter 2 - Inverse Trigonometric Functions - covers domains, ranges, properties, and simplification of inverse trig expressions. Carries 6-8 marks. Master the principal value branches and key properties.

Key Concepts

Principal Value Branches

Function	Domain	Range (Principal Value)
sin⁻¹x	[−1, 1]	[−π/2, π/2]
cos⁻¹x	[−1, 1]	[0, π]
tan⁻¹x	R (all reals)	(−π/2, π/2)
cot⁻¹x	R	(0, π)
sec⁻¹x	R − (−1, 1)	[0, π] − {π/2}
cosec⁻¹x	R − (−1, 1)	[−π/2, π/2] − {0}

Important Properties

Complementary:
sin⁻¹x + cos⁻¹x = π/2
tan⁻¹x + cot⁻¹x = π/2
sec⁻¹x + cosec⁻¹x = π/2

Negative arguments:
sin⁻¹(−x) = −sin⁻¹x
cos⁻¹(−x) = π − cos⁻¹x
tan⁻¹(−x) = −tan⁻¹x

Reciprocal:
sin⁻¹(1/x) = cosec⁻¹x
cos⁻¹(1/x) = sec⁻¹x
tan⁻¹(1/x) = cot⁻¹x (for x > 0)

Sum formulas:
tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)), if xy < 1
tan⁻¹x − tan⁻¹y = tan⁻¹((x−y)/(1+xy))

Double angle:
2tan⁻¹x = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²)) = tan⁻¹(2x/(1−x²))

Solved Examples

Example 1

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

Solution: sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6
(Range of sin⁻¹ is [−π/2, π/2], and sin(π/6) = 1/2)

Example 2

Q: Prove that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4

Solution: Using tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy))
= tan⁻¹((1/2 + 1/3)/(1 − 1/6)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹(1) = π/4 ✓

Example 3

Q: Simplify: cos⁻¹(1/2) + 2sin⁻¹(1/2)

Solution: cos⁻¹(1/2) = π/3; sin⁻¹(1/2) = π/6
= π/3 + 2(π/6) = π/3 + π/3 = 2π/3

Quick Revision Points

sin⁻¹x ∈ [−π/2, π/2]; cos⁻¹x ∈ [0, π]; tan⁻¹x ∈ (−π/2, π/2)
sin⁻¹x + cos⁻¹x = π/2 (always!)
sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x
tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) when xy < 1
2tan⁻¹x = sin⁻¹(2x/(1+x²)) for |x| ≤ 1
Always check if answer falls within principal value range!

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