📘 Trigonometry · Chapter MN18🎯 NDA Level : High Priority
Inverse trigonometric functions reverse the ordinary trig functions: sin−1x asks “what angle has sine equal to x?” The critical point is that trig functions are periodic (many angles give the same sine), so inverse functions must be restricted to a single principal value branch to give a unique output. For NDA, this chapter is tested through principal value evaluation, property-based simplifications, and equations.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding principal values: sin−1(1/2), cos−1(−1), tan−1(−√3), etc.; (2) Applying complementary pair: sin−1x + cos−1x = π/2; (3) Applying tan−1x + tan−1y formula (both addition and subtraction); (4) Simplifying expressions: sin−1(sinθ), tan(cos−1x), etc.; (5) Solving equations: 2 tan−1x = sin−1(...), tan−1x + tan−12x = π/4; (6) Converting between different inverse trig forms; (7) Domain/range restrictions and the principal value branch.
Topics at a Glance
① Principal Values
Domain, range, branches of sin⁻¹, cos⁻¹, tan⁻¹
② Standard Values
sin⁻¹(1/2)=π/6, cos⁻¹(0)=π/2, etc.
③ Complementary Pairs
sin⁻¹x+cos⁻¹x=π/2, tan⁻¹+cot⁻¹=π/2
④ Addition Formulas
tan⁻¹x+tan⁻¹y formula, conditions
⑤ Conversion Identities
sin⁻¹x=cos⁻¹(√(1−x²))=tan⁻¹(x/√(1−x²))
⑥ Equations & Simplification
sin(cos⁻¹x), tan(sin⁻¹x), solving equations
1. Principal Values & Branches
1.1
Why Principal Values? — Definition & Branches
Trig functions are many-to-one, so we restrict the range to make the inverse a function
Since sin(30°) = sin(150°) = sin(390°) = 1/2, asking “what is sin−1(1/2)?” has infinitely many answers. To get a unique answer, we restrict the range of the inverse to a chosen interval called the principal value branch.
⚡ Domain & Range (Principal Value Branch) of All Inverse Trig Functions
Function Domain (input) Range (principal value output)
sin⁻¹ x [−1, 1] [−π/2, π/2] (includes endpoints)
cos⁻¹ x [−1, 1] [0, π] (includes endpoints)
tan⁻¹ x (−∞, +∞) (−π/2, π/2) (open endpoints)
cot⁻¹ x (−∞, +∞) (0, π) (open endpoints)
sec⁻¹ x |x| ≥ 1 [0,π]\{π/2} (all of [0,π] except π/2)
cosec⁻¹ x |x| ≥ 1 [−π/2,π/2]\{0} (all except 0)
Key: sin⁻¹ and tan⁻¹ have ranges centred at 0 (positive and negative halves). cos⁻¹ range is [0, π] (only non-negative). This is the most commonly tested distinction in NDA.
Complete Domain & Range Reference
Function
Domain (x values)
Principal Value Range
Key Boundary Note
sin⁻¹ x
−1 ≤ x ≤ 1
−π/2 ≤ y ≤ π/2
Includes ±π/2
cos⁻¹ x
−1 ≤ x ≤ 1
0 ≤ y ≤ π
Includes 0 and π
tan⁻¹ x
all real numbers
−π/2 < y < π/2
Excludes ±π/2 (open)
cot⁻¹ x
all real numbers
0 < y < π
Excludes 0 and π (open)
sec⁻¹ x
x ≤ −1 or x ≥ 1
[0,π] \ {π/2}
Excludes π/2
cosec⁻¹ x
x ≤ −1 or x ≥ 1
[−π/2,π/2] \ {0}
Excludes 0
⚠ Critical Point for NDA:
sin−1(sin θ) = θ only if θ ∈ [−π/2, π/2]. If θ is outside this range, you must first reduce it.
cos−1(cos θ) = θ only if θ ∈ [0, π].
Example: sin−1(sin 5π/6) ≠ 5π/6 because 5π/6 is outside [−π/2, π/2].
sin(5π/6) = sin(π−5π/6) = sin(π/6). So sin−1(sin 5π/6) = π/6.
1.2
Principal Values at Standard Angles
These 18 values cover nearly every direct evaluation in NDA
5π/4 is outside [0, π]. cos(5π/4) = cos(π+π/4) = −cos(π/4) = −1/√2.
cos⁻¹(−1/√2) = π − cos⁻¹(1/√2) = π − π/4 = 3π/4. Never write cos⁻¹(cosθ) = θ unless θ ∈ [0,π]. Always reduce to the range first.
🤯 T2. Evaluate sin(cos⁻¹(3/5)).
Let cos⁻¹(3/5) = α. Then cosα = 3/5 and α ∈ [0,π/2] (since 3/5 > 0).
In this range, sin ≥ 0. sinα = √(1−9/25) = √(16/25) = 4/5.
sin(cos⁻¹(3/5)) = sinα = 4/5. Method: Let the expression equal α, extract the constraint on α from the inverse function, then compute the outer trig function using a right triangle.
2. Properties of Inverse Trigonometric Functions
2.1
Complementary Pairs & Negative Argument Rules
These six properties are tested directly as MCQs in NDA
P1 — Complementary Pair (sin/cos)
sin⁻¹ x + cos⁻¹ x = π/2
Valid for all x ∈ [−1, 1]. Implies cos⁻¹x = π/2 − sin⁻¹x.
P2 — Complementary Pair (tan/cot)
tan⁻¹ x + cot⁻¹ x = π/2
Valid for all real x. Implies cot⁻¹x = π/2 − tan⁻¹x.
P3 — Complementary Pair (sec/cosec)
sec⁻¹ x + cosec⁻¹ x = π/2
Valid for |x| ≥ 1.
P4 — Odd Functions (sin, tan)
sin⁻¹(−x) = −sin⁻¹(x) tan⁻¹(−x) = −tan⁻¹(x)
sin⁻¹ and tan⁻¹ are odd functions; graphs symmetric about origin.
P5 — cos⁻¹ (NOT odd)
cos⁻¹(−x) = π − cos⁻¹(x)
Range of cos⁻¹ is [0,π], so no negative outputs possible.
P6 — Reciprocal Relations
sin⁻¹(1/x) = cosec⁻¹(x) for |x|≥1 cos⁻¹(1/x) = sec⁻¹(x) for |x|≥1 tan⁻¹(1/x) = cot⁻¹(x) for x>0 = −π+cot⁻¹(x) for x<0
Be careful with the negative case for tan⁻¹(1/x)!
📌 The Most Tested Property in NDA — P1:
sin−1x + cos−1x = π/2 is used constantly. For example:
sin−1(1/2) + cos−1(1/2) = π/6 + π/3 = π/2 ✓
sin−1x = π/3 → cos−1x = π/2 − π/3 = π/6.
Any time you see sin−1 + cos−1 with the same argument, the answer is π/2 immediately.
2.2
tan⁻¹ Addition & Subtraction Formulas
The most important formula in this chapter — conditions determine which case applies
⚡ tan⁻¹ x + tan⁻¹ y Formula — Three Cases
ADDITION:
Case 1: xy < 1 (i.e., x>0,y>0 with xy<1, or one negative):
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)]
Case 2: x>0, y>0, and xy > 1 (both positive, product > 1):
tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x+y)/(1−xy)]
Case 3: x<0, y<0, and xy > 1 (both negative, |xy| > 1):
tan⁻¹x + tan⁻¹y = −π + tan⁻¹[(x+y)/(1−xy)]
SUBTRACTION (always one formula, no cases):
tan⁻¹x − tan⁻¹y = tan⁻¹[(x−y)/(1+xy)] (valid when xy > −1)
DOUBLE ANGLE:
2 tan⁻¹x = tan⁻¹[2x/(1−x²)] for |x| < 1
= sin⁻¹[2x/(1+x²)]
= cos⁻¹[(1−x²)/(1+x²)]
The π adjustment in Cases 2 and 3 is because the formula [(x+y)/(1−xy)] gives an angle outside the principal range (−π/2, π/2) when xy>1. Adding or subtracting π brings it back to the correct quadrant.
Using Case 1 (assuming xy<1): tan⁻¹[(x+2x)/(1−2x²)] = π/4.
⇒ (3x)/(1−2x²) = tan(π/4) = 1.
3x = 1−2x² ⇒ 2x²+3x−1 = 0.
x = [−3 ± √(9+8)]/4 = [−3 ± √17]/4.
Taking positive root (check domain): x = (−3+√17)/4 ≈ (−3+4.12)/4 ≈ 0.28. x = (−3+√17)/4.
Verify: 2x² = 2(17−6√17+9)/16 = (26−6√17)/8. Check xy = 2x² < 1 ✓.
Negative root x = (−3−√17)/4 < 0: check if tan⁻¹(x)+tan⁻¹(2x)=π/4 is satisfied (both negative → sum negative, not π/4). Rejected.
3. Conversion Between Inverse Trig Forms
3.1
Expressing sin⁻¹ in Terms of cos⁻¹, tan⁻¹, etc.
All inverse trig functions can be converted into each other using right-triangle geometry
⚡ Conversion Identities — Between All Forms
FROM sin⁻¹ x (x ∈ [0,1] for simplicity):
sin⁻¹x = cos⁻¹(√(1−x²))
= tan⁻¹(x/√(1−x²))
= cot⁻¹(√(1−x²)/x)
= sec⁻¹(1/√(1−x²))
= cosec⁻¹(1/x)
FROM tan⁻¹ x:
tan⁻¹x = sin⁻¹(x/√(1+x²))
= cos⁻¹(1/√(1+x²))
IMPORTANT SPECIAL CONVERSIONS:
2 tan⁻¹x = sin⁻¹(2x/(1+x²)) for |x| ≤ 1
2 tan⁻¹x = cos⁻¹((1−x²)/(1+x²)) for x ≥ 0
2 tan⁻¹x = tan⁻¹(2x/(1−x²)) for |x| < 1
Derivation method: Draw a right triangle. If sin⁻¹x = α, then sinα=x/1, so opposite=x, hypotenuse=1, adjacent=√(1−x²). Read off any other trig ratio of α directly from the triangle.
Let α=cos⁻¹x, β=cos⁻¹y, γ=cos⁻¹z. So α+β+γ=π.
cosγ = cos(π−α−β) = −cos(α+β).
z = −(cosαcosβ−sinαsinβ) = −xy+sinαsinβ.
⇒ z+xy = sinαsinβ = √(1−x²)√(1−y²).
Square both sides: (z+xy)² = (1−x²)(1−y²).
z²+2xyz+x²y² = 1−x²−y²+x²y².
z²+2xyz = 1−x²−y².
x²+y²+z²+2xyz = 1 ✓.
🤯 T7. Simplify: tan⁻¹[(cos x − sin x)/(cos x + sin x)].
Divide numerator and denominator by cos x:
= tan⁻¹[(1−tan x)/(1+tan x)].
Using tan(A−B) = (tanA−tanB)/(1+tanAtanB) with A=π/4 (tan=1), B=x:
(1−tanx)/(1+tanx) = tan(π/4−x).
= tan⁻¹[tan(π/4−x)] = π/4−x (provided π/4−x ∈ (−π/2, π/2)). This elegant simplification appears frequently in NDA as a direct application of compound angle identity recognition inside an inverse function.
📝 Master Formula Sheet — MN18 Inverse Trigonometric Functions
All critical formulae for rapid pre-exam revision.
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