ITF Formula Sheet

Domain & Rangeall 6 functions

Function	Domain	Range (Principal)	One-liner use
sin⁻¹(x)	\|x\| ≤ 1	[−π/2, π/2]	Smallest angle whose sin is x
cos⁻¹(x)	\|x\| ≤ 1	[0, π]	Non-negative angle whose cos is x
tan⁻¹(x)	x ∈ ℝ	(−π/2, π/2)	Defined for all reals — open interval
cot⁻¹(x)	x ∈ ℝ	(0, π)	Always gives angle in (0, π)
sec⁻¹(x)	\|x\| ≥ 1	[0, π/2) ∪ (π/2, π]	π/2 excluded from range
cosec⁻¹(x)	\|x\| ≥ 1	[−π/2, 0) ∪ (0, π/2]	0 excluded from range

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x > 0 → all 6 ITFs give acute angles

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x < 0 → sin⁻¹, tan⁻¹, cosec⁻¹ → IVth quadrant (−π/2 to 0)

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x < 0 → cos⁻¹, cot⁻¹, sec⁻¹ → IInd quadrant (obtuse)

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IIIrd quadrant NEVER used in range of any ITF

Property 1 — f(f⁻¹x) = xouter cancels inner · aperiodic

sin(sin⁻¹ x) = x

x ∈ [−1, 1] · Direct simplification

cos(cos⁻¹ x) = x

x ∈ [−1, 1] · Direct simplification

tan(tan⁻¹ x) = x

x ∈ ℝ · Most commonly used cancellation

cot(cot⁻¹ x) = x

x ∈ ℝ

sec(sec⁻¹ x) = x

|x| ≥ 1

cosec(cosec⁻¹ x) = x

|x| ≥ 1

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All 6 are aperiodic — no periodic behaviour

Property 2 — f⁻¹(f(x)) [piecewise, periodic]inner cancels outer conditionally

sin⁻¹(sin x) — period 2π, ODD

= xx ∈ [−π/2, π/2]

Principal branch — direct answer

= π − xx ∈ [π/2, π]

Reflect around π/2

= −π − xx ∈ [−π, −π/2]

Reflect around −π/2

cos⁻¹(cos x) — period 2π, EVEN

= xx ∈ [0, π]

Principal branch

= −xx ∈ [−π, 0]

Flip sign for negatives

tan⁻¹(tan x) — period π, ODD

= xx ∈ (−π/2, π/2)

Principal branch

= x − πx ∈ (π/2, 3π/2)

Subtract one period

= x + πx ∈ (−3π/2, −π/2)

Add one period

cot⁻¹(cot x) — period π, neither even nor odd

= xx ∈ (0, π)

Principal branch

= x + πx ∈ (−π, 0)

Add π for negatives

= x − πx ∈ (π, 2π)

Subtract π beyond π

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sec⁻¹(sec x) → period 2π, EVEN | cosec⁻¹(cosec x) → period 2π, ODD

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KEY TRICK: If x ∉ principal range, shift by period until it falls inside, then apply piecewise rule.

Property 3 — Complementary Pairs [Sum = π/2]convert between paired functions

sin⁻¹x + cos⁻¹x = π/2

|x| ≤ 1 · Convert sin⁻¹ ↔ cos⁻¹

tan⁻¹x + cot⁻¹x = π/2

x ∈ ℝ · Very frequently used in JEE

sec⁻¹x + cosec⁻¹x = π/2

|x| ≥ 1 · Less common but asked in Adv

Property 4 — Negative Argumentodd vs π-correction

ODD functions → negate

sin⁻¹(−x) = −sin⁻¹(x)

|x| ≤ 1 · Flip sign directly

tan⁻¹(−x) = −tan⁻¹(x)

x ∈ ℝ · Most commonly used

cosec⁻¹(−x) = −cosec⁻¹(x)

|x| ≥ 1

NOT odd — π correction

cos⁻¹(−x) = π − cos⁻¹(x)

|x| ≤ 1 · Adds π, does NOT negate

cot⁻¹(−x) = π − cot⁻¹(x)

x ∈ ℝ

sec⁻¹(−x) = π − sec⁻¹(x)

|x| ≥ 1

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Memory: sin, tan, cosec → ODD (negate) | cos, cot, sec → add π

Property 5 — Reciprocal Conversionscosec⁻¹→sin⁻¹ · sec⁻¹→cos⁻¹

cosec⁻¹(x) = sin⁻¹(1/x)

|x| ≥ 1 · Reduce cosec⁻¹ to sin⁻¹

sec⁻¹(x) = cos⁻¹(1/x)

|x| ≥ 1 · Reduce sec⁻¹ to cos⁻¹

cot⁻¹(x) = tan⁻¹(1/x)x > 0

Direct reciprocal

cot⁻¹(x) = π + tan⁻¹(1/x)x < 0

Negative x — add π correction

Property 6 — Addition & Subtraction Formulastan⁻¹ · sin⁻¹ · cos⁻¹

tan⁻¹ addition

tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)]x>0, y>0, xy < 1

Standard case · most used

tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x+y)/(1−xy)]x>0, y>0, xy > 1

Add π when xy > 1

tan⁻¹x + tan⁻¹y = π/2x>0, y>0, xy = 1 (reciprocals)

x and y are reciprocals

tan⁻¹x − tan⁻¹y = tan⁻¹[(x−y)/(1+xy)]x>0, y>0

Subtraction — denominator +xy

tan⁻¹x + tan⁻¹y + tan⁻¹z
= tan⁻¹[(x+y+z−xyz) / (1−xy−yz−zx)]x,y,z>0 & xy+yz+zx < 1

Three-term sum formula

sin⁻¹ addition

sin⁻¹x + sin⁻¹y = sin⁻¹[x√(1−y²) + y√(1−x²)]x>0, y>0, x²+y² ≤ 1

Both inside unit circle

sin⁻¹x + sin⁻¹y = π − sin⁻¹[x√(1−y²) + y√(1−x²)]x>0, y>0, x²+y² > 1

Outside unit circle

sin⁻¹x − sin⁻¹y = sin⁻¹[x√(1−y²) − y√(1−x²)]x>0, y>0

Subtraction form

cos⁻¹ addition

cos⁻¹x + cos⁻¹y = cos⁻¹[xy − √(1−x²)·√(1−y²)]x>0, y>0

Note the minus sign inside

cos⁻¹x − cos⁻¹y = cos⁻¹(xy + √(1−x²)·√(1−y²))x ≤ y, x,y > 0

Note PLUS sign — x ≤ y case

cos⁻¹x − cos⁻¹y = −cos⁻¹(xy + √(1−x²)·√(1−y²))x > y, x,y > 0

x > y case — negative

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For negative x or y — apply Property 4 first, then use formulas above

Simplified ITF — Double & Triple Angle Forms2tan⁻¹ substitution

sin⁻¹(2x / 1+x²)

= 2tan⁻¹x|x| ≤ 1

Standard — sub x = tanθ

= π − 2tan⁻¹xx > 1

Angle exceeded π/2

= −(π + 2tan⁻¹x)x < −1

Negative side

cos⁻¹((1−x²) / (1+x²))

= 2tan⁻¹xx ≥ 0

Non-negative x

= −2tan⁻¹xx < 0

Negate for negative x

tan⁻¹(2x / (1−x²))

= 2tan⁻¹x|x| < 1

Standard

= π + 2tan⁻¹xx < −1

= −(π − 2tan⁻¹x)x > 1

sin⁻¹(3x − 4x³) [Triple angle]

= 3sin⁻¹x|x| ≤ 1/2

Central range — direct triple

= π − 3sin⁻¹x1/2 ≤ x ≤ 1

Mirror from π

= −(π + 3sin⁻¹x)−1 ≤ x ≤ −1/2

Left side

cos⁻¹(4x³ − 3x) [Triple angle]

= 3cos⁻¹x1/2 ≤ x ≤ 1

= 2π − 3cos⁻¹x−1 ≤ x ≤ −1/2

= 2π − 3cos⁻¹x−1/2 ≤ x ≤ 1/2

Middle zone

sin⁻¹(2x√(1−x²)) [Double angle]

= 2sin⁻¹x|x| ≤ 1/√2

Sub x = sinθ

= π − 2sin⁻¹x1/√2 ≤ x ≤ 1

= −(π + 2sin⁻¹x)−1 ≤ x ≤ −1/√2

cos⁻¹(2x²−1) [Double angle]

= 2cos⁻¹x0 ≤ x ≤ 1

Sub x = cosθ

= 2π − 2cos⁻¹x−1 ≤ x ≤ 0

Half-Angle ITF Identitiesθ/2 forms

cos⁻¹√((1+x)/2) = sin⁻¹√((1−x)/2)

∀ x ∈ [−1,1] · via cos(θ/2) identity

cos⁻¹√((1+x)/2) = (1/2)cos⁻¹x

Half-angle compression

sin⁻¹√((1−x)/2) = (1/2)cos⁻¹x

Alternative form of same identity

tan⁻¹√x = (1/2)cos⁻¹((1−x)/(1+x))x ∈ [0, 1]

Useful in domain-range finding problems

Equation Solving Tricksquick shortcuts

sin⁻¹f(x) + cos⁻¹g(x) = π/2 ⇒ f(x) = g(x)

Then check domain |f|, |g| ≤ 1

sin⁻¹(f) + cos⁻¹(f) = π/2 always

Direct use of Property 3

2tan⁻¹x = sin⁻¹(2x/(1+x²))|x| ≤ 1

Convert 2tan⁻¹ to single ITF

tan⁻¹(tan x) = x − nπshift x into (−π/2, π/2)

Evaluate out-of-range tan⁻¹(tan x)

cot⁻¹(cot x) = x − nπshift x into (0, π)

Evaluate cot⁻¹(cot x)

cos⁻¹(cos x) = 2nπ ± xbring into [0, π]

Reduce using symmetry of cosine

Inequation Rules — Monotonicitydirection of inequality

x > y ⇔ sin⁻¹x > sin⁻¹y

|x|,|y| ≤ 1 · INCREASING — preserved

x > y ⇔ cos⁻¹x < cos⁻¹y

|x|,|y| ≤ 1 · DECREASING — FLIPS

x > y ⇔ tan⁻¹x > tan⁻¹y

x,y ∈ ℝ · INCREASING — preserved

x > y ⇔ cot⁻¹x < cot⁻¹y

x,y ∈ ℝ · DECREASING — FLIPS

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cos⁻¹ and cot⁻¹ are decreasing → flip the inequality sign when comparing values

Series Summation — Telescoping via tan⁻¹tan⁻¹x − tan⁻¹y form

tan⁻¹[(x−y)/(1+xy)] = tan⁻¹x − tan⁻¹y

Core identity — split into telescoping difference

Σ tan⁻¹(1/(n²−n+1))T_n = tan⁻¹(n) − tan⁻¹(n−1)

Telescopes to S∞ = π/2

Σ tan⁻¹(2/n²)T_n = tan⁻¹(n+1) − tan⁻¹(n−1)

Two-step telescope — careful boundary

Σ cot⁻¹(2ⁿ⁺¹ + 2⁻ⁿ) = π/4

Convert cot⁻¹ → tan⁻¹, then telescope

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TRICK: Write T_n = tan⁻¹(f(n)) − tan⁻¹(f(n−1)) → S_n = tan⁻¹(f(n)) − tan⁻¹(f(0))

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To split: express numerator as A − B, denominator as 1 + AB, match tan⁻¹A − tan⁻¹B form

Standard Values — Must Knowmemorise all

sin⁻¹(0)= 0

sin⁻¹(1/2)= π/6

sin⁻¹(1/√2)= π/4

sin⁻¹(√3/2)= π/3

sin⁻¹(1)= π/2

sin⁻¹(−1/2)= −π/6

sin⁻¹(−1/√2)= −π/4

sin⁻¹(−1)= −π/2

cos⁻¹(1)= 0

cos⁻¹(√3/2)= π/6

cos⁻¹(1/√2)= π/4

cos⁻¹(1/2)= π/3

cos⁻¹(0)= π/2

cos⁻¹(−1/2)= 2π/3

cos⁻¹(−1/√2)= 3π/4

cos⁻¹(−1)= π

tan⁻¹(0)= 0

tan⁻¹(1/√3)= π/6

tan⁻¹(1)= π/4

tan⁻¹(√3)= π/3

tan⁻¹(∞)= π/2

sec⁻¹(1)= 0

sec⁻¹(√2)= π/4

sec⁻¹(2)= π/3

Cross-Conversion — sin/cos/tan of Inverse Functionsright triangle method

sin(cos⁻¹ x)	= √(1−x²)
cos(sin⁻¹ x)	= √(1−x²)
tan(sin⁻¹ x)	= x / √(1−x²)
tan(cos⁻¹ x)	= √(1−x²) / x
sin(tan⁻¹ x)	= x / √(1+x²)

cos(tan⁻¹ x)	= 1 / √(1+x²)
sec(tan⁻¹ x)	= √(1+x²)
cosec(cot⁻¹ x)	= √(1+x²)
sin⁻¹(a/b)	= tan⁻¹(a / √(b²−a²))
cos⁻¹(a/b)	= tan⁻¹(√(b²−a²) / a)

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Method: Set f⁻¹(x) = θ, draw right triangle, use Pythagoras for the needed trig ratio

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Classic: sec²(tan⁻¹2) + cosec²(cot⁻¹3) = (1+4) + (1+9) = 15

Inverse Trig Functions