NCERT Plus Two (Class 12) Mathematics • Chapter 2

Inverse Trigonometric Functions

Q: Find the principal value of cosec⁻¹(2) and tan⁻¹(−√3).

• cosec⁻¹(2) = sin⁻¹(1/2) = π/6 (∈ [−π/2, π/2] − {0}). • tan⁻¹(−√3) = −tan⁻¹(√3) = −π/3 (∈ (−π/2, π/2)).

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

Note 7π/6 ∉ [0, π] ! Rewrite cos(7π/6) = cos(2π − 5π/6) = cos(5π/6) . Since 5π/6 ∈ [0, π] , answer = 5π/6 .

Q: Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).

tan⁻¹(1) = π/4 cos⁻¹(−1/2) = π − π/3 = 2π/3 sin⁻¹(−1/2) = −π/6 Sum = π/4 + 2π/3 − π/6 = 3π/4 .

Q: If sin(sin⁻¹(1/5) + cos⁻¹ x) = 1, find the value of x.

Since sin(π/2) = 1 , we get sin⁻¹(1/5) + cos⁻¹ x = π/2 . Using identity sin⁻¹(a) + cos⁻¹(a) = π/2 for a ∈ [−1, 1] , we get x = 1/5 .

Q: Write tan⁻¹( √((1 − cos x)/(1 + cos x)) ), 0 < x < π, in the simplest form.

Use half-angle formulas: 1 − cos x = 2 sin²(x/2) and 1 + cos x = 2 cos²(x/2) . Inside radical: √(tan²(x/2)) = tan(x/2) (positive since 0 < x/2 < π/2 ). Hence tan⁻¹(tan(x/2)) = x/2 .

Q: Express tan⁻¹( (cos x − sin x)/(cos x + sin x) ), −π/4 < x < 3π/4, in simplest form.

Divide numerator and denominator by cos x : = tan⁻¹( (1 − tan x)/(1 + tan x) ) = tan⁻¹(tan(π/4 − x)) . Since −π/4 < x < 3π/4 ⇒ −π/2 < π/4 − x < π/2 , answer = π/4 − x .

Q: Prove that 3 sin⁻¹ x = sin⁻¹(3x − 4x³), x ∈ [−1/2, 1/2].

Let sin⁻¹ x = θ ⇒ x = sin θ with θ ∈ [−π/6, π/6] (so 3θ ∈ [−π/2, π/2] ). RHS = sin⁻¹(3 sin θ − 4 sin³θ) = sin⁻¹(sin 3θ) = 3θ = 3 sin⁻¹ x = LHS .

Q: Write in simplest form: (a) tan⁻¹( x / √(a² − x²) ), |x| < a (b) tan⁻¹( √(1 + x²) − 1 ) / x, x ≠ 0.

• (a) Put x = a sin θ ⇒ tan⁻¹(a sin θ / a cos θ) = tan⁻¹(tan θ) = θ = sin⁻¹(x/a) . • (b) Put x = tan θ ⇒ (sec θ − 1)/tan θ = (1 − cos θ)/sin θ = tan(θ/2) ⇒ θ/2 = ½ tan⁻¹ x .

Domain restrictions, principal value branches [-π/2, π/2] vs [0, π], graphs & simplification identities.

📄 Official NCERT Textbook PDF

Topic 2.2: Why Trigonometric Domains Must Be Restricted for Inverses

Ask your class: "Which angle has sin θ = 1/2?" Arjun says 30° (π/6), Maya says 150° (5π/6), and Kiran yells 390°! Because trig waves repeat infinitely, a single output 1/2 comes from infinitely many angles — failing Chapter 1's One-One test! To unlock a genuine inverse sin⁻¹x, mathematicians fence off ONE neat angle window around zero where every height happens exactly once!

THE PROBLEMFull trig functions (sin: ℝ → [−1, 1]) are Many-One, so they have NO inverse!

THE FIXRestrict domain to one monotonic slice: [−π/2, π/2] for sin, [0, π] for cos!

MEANING OF sin⁻¹xsin⁻¹x returns an ANGLE θ whose sine is x (NEVER 1/sin x!)

🧠 Memory Trick: Inverse Trig = Angle Finder! Fence off the piece closest to 0 that climbs smoothly from −1 to +1 without ever turning back!

💡 Simple Kid-Check: In Chapter 1, you learned only Bijective (One-One + Onto) functions have an UNDO button! A roller coaster going up and down fails the Horizontal Line Test. By chopping the wave to a single smooth hill or slide, every output now traces back to ONE unique angle!

1
Why [−π/2, π/2] works for sin x: On [−π/2, π/2], sin x climbs strictly from −1 to +1 — hitting every value in [−1, 1] exactly once!
2
Why [−π/2, π/2] FAILS for cos x: Because cos(−60°) = cos(+60°) = 1/2, it repeats! So for cos x, we shift to [0, π], where it falls strictly from +1 down to −1!
3
What about tan x? Restricting to open (−π/2, π/2) avoids the vertical walls at ±π/2 and covers all real numbers (−∞, ∞) smoothly!

Biggest Chapter 2 beginner trap: thinking sin⁻¹ x means (sin x)⁻¹ = 1 / sin x! NO! 1 / sin x is written (sin x)⁻¹ or cosec x, whereas sin⁻¹ x (also called arcsin x) is the inverse function that outputs an ANGLE!

Topic 2.2: Principal Value Branches & Graphs of arcsin, arccos & arctan

Take the restricted slice of sin x, hold a mirror right along the diagonal line y = x, and flip the curve over! Inputs and outputs swap places instantly: numbers x ∈ [−1, 1] now go IN, and angles inside [−π/2, π/2] come OUT! That chosen official output range is called the Principal Value Branch!

sin⁻¹ x (arcsin)Domain: [−1, 1]
Branch: [−π/2, π/2] (Increasing ↗)

cos⁻¹ x (arccos)Domain: [−1, 1]
Branch: [0, π] (Decreasing ↘)

tan⁻¹ x (arctan)Domain: ℝ (−∞, ∞)
Branch: (−π/2, π/2) (Open ends!)

🧠 Memory Trick: sin⁻¹ & tan⁻¹ live on the RIGHT semicircle (−π/2 to +π/2); cos⁻¹ lives on the TOP semicircle (0 to π)!

💡 Simple Kid-Check: Notice the graph flip across y = x! Since sin x is squashed between heights −1 and +1 on width [−π/2, π/2], its mirror twin sin⁻¹ x sits on width [−1, 1] and stretches vertically from −π/2 to +π/2!

Function	Domain (Input x)	Principal Branch (Output y)	Key Values
`y = sin⁻¹ x`	`[−1, 1]`	`[−π/2, π/2]` (Closed)	`sin⁻¹(1) = π/2`, `sin⁻¹(0) = 0`
`y = cos⁻¹ x`	`[−1, 1]`	`[0, π]` (Closed)	`cos⁻¹(−1) = π`, `cos⁻¹(1) = 0`
`y = tan⁻¹ x`	`ℝ`	`(−π/2, π/2)` (Open)	`tan⁻¹(1) = π/4`, asymptotes `±π/2`

🧒 Quick Principal Value Check: Find the principal value of sin⁻¹(sin(2π3)).
Does 2π/3 sit inside [−π/2, π/2]? NO! (120° > 90°).
Rewrite sin(2π3) = sin(π − π3) = sin(π3) → Now π/3 ∈ [−π/2, π/2] ⇒ Answer = π/3!

NEVER blindly cancel sin⁻¹(sin θ) = θ! That shortcut works ONLY when θ ∈ [−π/2, π/2]! For sin⁻¹(sin(2π/3)), writing 2π/3 loses full marks — always fold θ back into the principal branch first! Also notice tan⁻¹ x uses OPEN parentheses (−π/2, π/2) because ±π/2 are horizontal asymptotes never touched!

Topic 2.2: Principal Values of cosec⁻¹x, sec⁻¹x & cot⁻¹x

What about the three reciprocal cousins: cosec⁻¹x, sec⁻¹x, and cot⁻¹x? Think of them as moving into the exact same two neighbourhoods — cosec⁻¹ rooms with sin⁻¹, while sec⁻¹ and cot⁻¹ live upstairs with cos⁻¹! They only drill one tiny safety hole wherever dividing by zero would crash!

cosec⁻¹ xDomain: ℝ − (−1, 1) (|x| ≥ 1)
Branch: [−π/2, π/2] − {0}

sec⁻¹ xDomain: ℝ − (−1, 1) (|x| ≥ 1)
Branch: [0, π] − {π/2}

cot⁻¹ xDomain: ℝ (−∞, ∞)
Branch: (0, π) (Open!)

🧠 Memory Trick: TWO TEAMS!
• Right-Half Team (−π/2 to π/2): sin⁻¹, tan⁻¹, cosec⁻¹
• Top-Half Team (0 to π): cos⁻¹, sec⁻¹, cot⁻¹!

💡 Simple Kid-Check: Why is 0 kicked out of cosec⁻¹, and π/2 kicked out of sec⁻¹? Because sin 0 = 0 (so cosec 0 = 1/0 blows up!) and cos(π/2) = 0 (so sec(π/2) = 1/0 blows up!). And for cot⁻¹, both ends 0 and π make sin = 0, leaving open (0, π)!

Function	Domain	Principal Value Branch	Easy Evaluation Link
`cosec⁻¹ x`	`(−∞, −1] ∪ [1, ∞)`	`[−π/2, π/2] − {0}`	`cosec⁻¹(2) = sin⁻¹(1/2) = π/6`
`sec⁻¹ x`	`(−∞, −1] ∪ [1, ∞)`	`[0, π] − {π/2}`	`sec⁻¹(2/√3) = cos⁻¹(√3/2) = π/6`
`cot⁻¹ x`	`ℝ`	`(0, π)`	`cot⁻¹(√3) = tan⁻¹(1/√3) = π/6` (for `x > 0`)

HUGE Kerala Board MCQ trap: placing cot⁻¹ x in (−π/2, π/2) just like tan⁻¹ x! WRONG! In NCERT, cot⁻¹ x lives strictly in (0, π) — alongside cos⁻¹ and sec⁻¹! Always remember: C-S-C (Cos, Sec, Cot) = Top Half [0, π]! Also never forget −{0} for cosec⁻¹ and −{π/2} for sec⁻¹!

Topic 2.2: Evaluating Negative-Argument Principal Values (arccos(−x) Trap!)

What happens when you drop a negative number like −1/2 into an inverse trig machine? Team Right-Half (sin⁻¹, tan⁻¹, cosec⁻¹) simply steps downstairs into Quadrant IV, so the minus slips straight outside: sin⁻¹(−1/2) = −π/6! Try that with cos⁻¹(−1/2) and you hit a locked door — negative angles are banned in [0, π]! So Team Top-Half steps left into Quadrant II: π − cos⁻¹(1/2) = π − π/3 = 2π/3!

TEAM RIGHT-HALF (− OUT)sin⁻¹(−x) = −sin⁻¹x
tan⁻¹(−x) = −tan⁻¹x
cosec⁻¹(−x) = −cosec⁻¹x

TEAM TOP-HALF (π − RULE!)cos⁻¹(−x) = π − cos⁻¹x
sec⁻¹(−x) = π − sec⁻¹x
cot⁻¹(−x) = π − cot⁻¹x

WHY π − ?cos(π − θ) = −cos θ puts the angle neatly inside Quadrant II (π/2, π]!

🧠 Memory Trick: Right-Half spits minus OUT (−θ); Top-Half subtracts from π (π − θ)! Never put a negative angle inside cos⁻¹, sec⁻¹, or cot⁻¹!

💡 Simple Kid-Check: Picture a dial! If your angle needle lives in [−π/2, π/2], negative inputs flip the needle down into (−π/2, 0). If your needle lives in [0, π], it can NEVER dip below 0 — so negative inputs swing it to the obtuse side between π/2 and π via π − θ!

🧒 Classic 2-Mark Board Problem: Find tan⁻¹(1) + cos⁻¹(−12) + sin⁻¹(−12).
• tan⁻¹(1) = π4
• cos⁻¹(−12) = π − cos⁻¹(12) = π − π3 = 2π3
• sin⁻¹(−12) = −sin⁻¹(12) = −π6
Total = π4 + 2π3 − π6 = 3π + 8π − 2π12 = 9π12 = 3π4!

The #1 most common 1-mark mistake in Kerala HSE exams: writing cos⁻¹(−1/2) = −π/3 or cot⁻¹(−√3) = −π/6! Both are WRONG! Angles in cos⁻¹, sec⁻¹, cot⁻¹ can NEVER be negative! Correct answers: cos⁻¹(−1/2) = π − π/3 = 2π/3, and cot⁻¹(−√3) = π − π/6 = 5π/6!

Topic 2.3: Core Simplification Identities (sin⁻¹(1/x), sin⁻¹x + cos⁻¹x = π/2)

Draw a right triangle with hypotenuse 1 and opposite side x. One acute corner angle has sin α = x (so α = sin⁻¹x). Look at the OTHER acute corner β: that same side x is ADJACENT to β, so cos β = x (β = cos⁻¹x)! Since the two acute corners of a right triangle ALWAYS add up to 90°, sin⁻¹x + cos⁻¹x = π/2 is pure triangle magic!

COMPLEMENTARY (π/2)sin⁻¹x + cos⁻¹x = π/2 (x∈[−1,1])
tan⁻¹x + cot⁻¹x = π/2 (x∈ℝ)
sec⁻¹x + cosec⁻¹x = π/2 (|x|≥1)

RECIPROCAL IDENTITIESsin⁻¹(1/x) = cosec⁻¹x (|x|≥1)
cos⁻¹(1/x) = sec⁻¹x (|x|≥1)
tan⁻¹(1/x) = cot⁻¹x (x > 0)

TRIANGLE CONVERTERPut sin⁻¹(a/c) = θ ⇒ build triangle with Opp=a, Hyp=c, Adj=√(c²−a²)!

🧠 Memory Trick: Co-functions sum to π/2 (sin + cos, tan + cot, sec + cosec = π/2)! And check domain x > 0 before flipping tan⁻¹(1/x) = cot⁻¹x!

💡 Simple Kid-Check: Whenever you need to convert sin⁻¹ into cos⁻¹ or tan⁻¹, never memorize messy formulas — just sketch a tiny right triangle with sides from Pythagoras and read off the ratio!

Triple-Angle Classics (NCERT Proofs):
• 3 sin⁻¹ x = sin⁻¹(3x − 4x³) for x ∈ [−1/2, 1/2] (Put x = sin θ!)
• 3 cos⁻¹ x = cos⁻¹(4x³ − 3x) for x ∈ [1/2, 1] (Put x = cos θ!)

🧒 Quick 2-Mark Board Trick: If sin(sin⁻¹(15) + cos⁻¹ x) = 1, find x!
Since sin(π/2) = 1, we need sin⁻¹(15) + cos⁻¹ x = π2.
Compare directly with sin⁻¹(15) + cos⁻¹(15) = π2 ⇒ x = 15 in 2 lines!

Watch TWO domain traps! (1) sin⁻¹ x + cos⁻¹ x = π/2 works ONLY for x ∈ [−1, 1] — never plug x = 3! (2) tan⁻¹(1/x) = cot⁻¹ x holds ONLY for x > 0! When x < 0, tan⁻¹(1/x) is negative while cot⁻¹ x is in (π/2, π), so you get tan⁻¹(1/x) = −π + cot⁻¹ x!

Topic 2.3: Algebraic Simplification Using Trigonometric Substitutions

Staring at cot⁻¹(1 / √(x² − 1)) feels like staring at a tangled coconut rope! How do you melt square roots away in two seconds? Slip on the secret trig disguise x = sec θ! Instantly √(sec²θ − 1) transforms into tan θ, the fraction flips to cot θ, and the whole scary expression collapses to θ = sec⁻¹x!

√(a² − x²) DISGUISEPut x = a sin θ or a cos θ
(Uses 1 − sin²θ = cos²θ)

√(a² + x²) DISGUISEPut x = a tan θ or a cot θ
(Uses 1 + tan²θ = sec²θ)

√(x² − a²) DISGUISEPut x = a sec θ or a cosec θ
(Uses sec²θ − 1 = tan²θ)

🧠 Memory Trick: Minus x² → sin; Plus x² → tan; x² first (minus a²) → sec! And for (1 − cos x)/(1 + cos x), always fire half-angle 2 sin²(x/2) / 2 cos²(x/2)!

💡 Simple Kid-Check: Radical expressions hate square roots! Each trigonometric substitution matches one of the three Plus One Pythagorean identities (1 − sin²θ, 1 + tan²θ, sec²θ − 1) so the inside turns into a neat perfect square!

Expression Inside	Best Substitution	Why It Collapses
`√(a² − x²)`	`x = a sin θ` or `a cos θ`	`√(a²(1 − sin²θ)) = a cos θ`
`√(a² + x²)`	`x = a tan θ` or `a cot θ`	`√(a²(1 + tan²θ)) = a sec θ`
`√(x² − a²)`	`x = a sec θ` or `a cosec θ`	`√(a²(sec²θ − 1)) = a tan θ`
`√((a − x)/(a + x))`	`x = a cos 2θ`	`√(tan²θ) = tan θ`

🧒 Two Must-Know 3-Mark NCERT Simplifications:
1. Simplify tan⁻¹( cos x − sin xcos x + sin x ), −π/4 < x < 3π/4.
Divide numerator & denominator by cos x:
= tan⁻¹( 1 − tan x1 + tan x ) = tan⁻¹(tan(π4 − x)) = π4 − x!

2. Simplify tan⁻¹( x√(a² − x²) ), |x| < a.
Put x = a sin θ (θ ∈ (−π/2, π/2)): a sin θa cos θ = tan θ ⇒ θ = sin⁻¹(xa)!

Watch TWO sneaky simplification traps! (1) When taking √(tan²θ) or √(cos²θ), check the domain given in the question so you know whether it's + or −! (2) After simplifying to tan⁻¹(tan θ), double-check that θ ∈ (−π/2, π/2) before dropping tan⁻¹(tan) — if x = a sec θ has x < −1, θ lies in (π/2, π] and needs adjustment!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find the principal value of cosec⁻¹(2) and tan⁻¹(−√3).

• cosec⁻¹(2) = sin⁻¹(1/2) = π/6 (∈ [−π/2, π/2] − {0}).
• tan⁻¹(−√3) = −tan⁻¹(√3) = −π/3 (∈ (−π/2, π/2)).

Q2 (Kerala HSE Say • 1 Marks): Find the value of cos⁻¹(cos(7π/6)).

Note 7π/6 ∉ [0, π]!
Rewrite cos(7π/6) = cos(2π − 5π/6) = cos(5π/6).
Since 5π/6 ∈ [0, π], answer = 5π/6.

Q3 (Kerala HSE March • 2 Marks): Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).

tan⁻¹(1) = π/4
cos⁻¹(−1/2) = π − π/3 = 2π/3
sin⁻¹(−1/2) = −π/6
Sum = π/4 + 2π/3 − π/6 = 3π/4.

Q4 (NCERT / DHSE Model • 2 Marks): If sin(sin⁻¹(1/5) + cos⁻¹ x) = 1, find the value of x.

Since sin(π/2) = 1, we get sin⁻¹(1/5) + cos⁻¹ x = π/2.
Using identity sin⁻¹(a) + cos⁻¹(a) = π/2 for a ∈ [−1, 1], we get x = 1/5.

Q5 (Kerala HSE March • 3 Marks): Write tan⁻¹( √((1 − cos x)/(1 + cos x)) ), 0 < x < π, in the simplest form.

Use half-angle formulas: 1 − cos x = 2 sin²(x/2) and 1 + cos x = 2 cos²(x/2).
Inside radical: √(tan²(x/2)) = tan(x/2) (positive since 0 < x/2 < π/2).
Hence tan⁻¹(tan(x/2)) = x/2.

Q6 (Kerala HSE Say • 3 Marks): Express tan⁻¹( (cos x − sin x)/(cos x + sin x) ), −π/4 < x < 3π/4, in simplest form.

Divide numerator and denominator by cos x:
= tan⁻¹( (1 − tan x)/(1 + tan x) ) = tan⁻¹(tan(π/4 − x)).
Since −π/4 < x < 3π/4 ⇒ −π/2 < π/4 − x < π/2, answer = π/4 − x.

Q7 (Kerala HSE March • 3 Marks): Prove that 3 sin⁻¹ x = sin⁻¹(3x − 4x³), x ∈ [−1/2, 1/2].

Let sin⁻¹ x = θ ⇒ x = sin θ with θ ∈ [−π/6, π/6] (so 3θ ∈ [−π/2, π/2]).
RHS = sin⁻¹(3 sin θ − 4 sin³θ) = sin⁻¹(sin 3θ) = 3θ = 3 sin⁻¹ x = LHS.

Q8 (Kerala HSE Model • 4 Marks): Write in simplest form: (a) tan⁻¹( x / √(a² − x²) ), |x| < a (b) tan⁻¹( √(1 + x²) − 1 ) / x, x ≠ 0.

• (a) Put x = a sin θ ⇒ tan⁻¹(a sin θ / a cos θ) = tan⁻¹(tan θ) = θ = sin⁻¹(x/a).
• (b) Put x = tan θ ⇒ (sec θ − 1)/tan θ = (1 − cos θ)/sin θ = tan(θ/2) ⇒ θ/2 = ½ tan⁻¹ x.