Chapter 2 of 13

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CBSE Class 12 Mathematics★ Unit I · 8 marks, shared with Relations and Functions

Inverse Trigonometric Functions

Trig functions repeat, so they aren't one-one on their full domain — restrict each one to an interval where it is, and you get an inverse with a well-defined principal value branch. The current book teaches every simplification through one substitution technique (x=sinθ, x=cosθ, x=tanθ, x=secθ) rather than a list of identities to memorise. CBSE Class 12 Mathematics Chapter 2.

⏱ 9 min readMedium difficulty✓ Free · NCERT-aligned

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1State the domain and principal value range of all six inverse trigonometric functions
2Find the principal value of an inverse trig function applied to a standard ratio (like 1/2, -1, sqrt3)
3Distinguish sin^{-1}x from (sin x)^{-1} and apply sin(sin^{-1}x)=x versus sin^{-1}(sin x)=x correctly, including when x falls outside the principal branch
4Simplify a composite inverse-trig expression by substituting x as sinθ, cosθ, tanθ, or secθ and reducing with a standard trig identity
5Solve equations involving inverse trig functions and check candidate solutions against the original equation to eliminate extraneous roots

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Why this chapter matters

Every derivative and integral of an inverse trig function later in Class 12 depends on knowing exactly which interval its principal value lives in — get the branch wrong here and a correct-looking calculus answer comes out with the wrong sign or the wrong angle.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Trigonometric functions (Class 11): domains, ranges, and periodicity of sin, cos, tan, cot, sec, cosec, and their graphs

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Invertibility (Class 12, Relations and Functions): a function is invertible if and only if it is bijective, which is exactly why trig functions need their domains restricted before an inverse can be defined

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Trigonometric identities (Class 11): double angle, half angle, and triple angle formulas, used throughout every substitution-based simplification in this chapter

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Relations and Functions (Class 12 Ch 1): a function is invertible exactly when it is bijective, which is precisely why each trig function's domain must be restricted before an inverse exists

Inverse Trigonometric Functions

1. Check this before you revise anything

There is no boxed list of "properties" to memorise in the current book. Older editions of this chapter had a numbered list of identities — sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2, the tan⁻¹ addition formula, and so on. The current (2026-27) edition drops that list entirely: Section 2.3 goes straight from a one-line recap of sin(sin⁻¹x)=x into three worked examples.

Every one of those three examples solves a simplification by substituting x=sinθ, x=cosθ, x=tanθ, or x=secθ and then applying an ordinary trig identity (double angle, half angle, or triple angle). The skill being tested is the substitution technique itself, not formula recall — and every real exercise question follows the same pattern.

The old stub taught that removed identity list as if it were still boxed content, and its ncertExercises field collapsed the chapter's three real problem sets into one invented group of 21 questions with no solutions file behind it at all. The book actually has Exercise 2.1 (14 questions, principal values), Exercise 2.2 (15 questions, proving identities and simplifying expressions by substitution), and a Miscellaneous Exercise (14 questions) — 43 questions in total.

2. What this chapter covers

Textbook section	Topic
2.2	Restricting each trig function's domain to get a one-one, onto (invertible) piece; principal value branches
2.3	Simplifying composite inverse-trig expressions via substitution

3. Principal value branches

A trig function isn't invertible on its natural domain because it repeats — sin(0)=sin(π)=0, for instance. Restrict the domain to one interval where the function is one-one and onto, and the inverse becomes well defined on that piece. The interval CBSE calls the principal value branch is the standard choice:

Function	Domain	Principal value range
$sin^{- 1} x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$cos^{- 1} x$	$[- 1, 1]$	$[0, π]$
$cosec^{- 1} x$	$R - (- 1, 1)$	$[- \frac{π}{2}, \frac{π}{2}] - {0}$
$sec^{- 1} x$	$R - (- 1, 1)$	$[0, π] - {\frac{π}{2}}$
$tan^{- 1} x$	$R$	$(- \frac{π}{2}, \frac{π}{2})$
$cot^{- 1} x$	$R$	$(0, π)$

The book's own note worth remembering: $sin^{- 1} x$ is not the same as $(sin x)^{- 1}$ . The second is just $1/ sin x$ ; the raised $- 1$ on an inverse trig function always means "the inverse function," never "reciprocal."

Why the branches are chosen the way they are. Each range is the shortest interval on which the function runs monotonically through its entire set of values exactly once. For sine that is $[- \frac{π}{2}, \frac{π}{2}]$ , where it climbs steadily from $- 1$ to $1$ . For cosine the same requirement forces a different interval, $[0, π]$ , where it falls from $1$ to $- 1$ — which is why the sine and cosine branches do not match.

The gaps in the cosec and sec ranges exist because those functions are undefined where their reciprocals vanish. $cosec^{- 1}$ excludes $0$ from its range since $cosec θ$ is undefined at $θ = 0$ , and $sec^{- 1}$ excludes $\frac{π}{2}$ for the same reason.

Open versus closed intervals matter. $tan^{- 1}$ and $cot^{- 1}$ have open ranges, because $tan θ$ and $cot θ$ run off to infinity at the endpoints and never actually attain a value there. Writing $[- \frac{π}{2}, \frac{π}{2}]$ for $tan^{- 1}$ is a marked error.

The two composition rules are not symmetric, and this is the single most examined subtlety in the chapter:

$sin (sin^{- 1} x) = x for all x \in [- 1, 1]$

$sin^{- 1} (sin x) = x only when x \in [- \frac{π}{2}, \frac{π}{2}]$

The first always holds because $sin^{- 1} x$ lands inside the branch by construction. The second fails whenever $x$ starts outside the branch.

When $x$ lies outside the branch, replace it by the angle inside the branch having the same sine. For $sin^{- 1} (sin \frac{3 π}{4})$ , note $sin \frac{3 π}{4} = sin (π - \frac{3 π}{4}) = sin \frac{π}{4}$ , and $\frac{π}{4}$ is in the branch, so the answer is $\frac{π}{4}$ — not $\frac{3 π}{4}$ .

Negative arguments. For the branches symmetric about the origin, the inverse is an odd function: $sin^{- 1} (- x) = - sin^{- 1} x$ and $tan^{- 1} (- x) = - tan^{- 1} x$ . For $cos^{- 1}$ and $cot^{- 1}$ , whose ranges sit in $[0, π]$ , the rule is different: $cos^{- 1} (- x) = π - cos^{- 1} x$ . Applying the odd-function rule to $cos^{- 1}$ is a common and costly slip.

Worked, mirroring the textbook's own Example 2. Find the principal value of $cot^{- 1} (\frac{- 1}{3})$ . Let $y = cot^{- 1} (\frac{- 1}{3})$ , so $cot y = \frac{- 1}{3} = - cot \frac{π}{3} = cot (π - \frac{π}{3}) = cot \frac{2 π}{3}$ . Since the principal branch of $cot^{- 1}$ is $(0, π)$ and $\frac{2 π}{3}$ lies in it, the principal value is $\frac{2 π}{3}$ .

4. The substitution technique

Every simplification in this chapter follows the same shape: spot which trig ratio the expression looks like, substitute $x$ as that ratio of a new angle $θ$ , collapse the expression using a standard identity, then read off the answer as a multiple of $θ$ — always tracking which interval $θ$ (and any multiple of it) must lie in for the principal branch to apply.

Worked, mirroring the textbook's own Example 3(i). Show $sin^{- 1} (2 x 1 - x^{2}) = 2 sin^{- 1} x$ for $- \frac{1}{2} \leq x \leq \frac{1}{2}$ . Let $x = sin θ$ , so $sin^{- 1} x = θ$ . Then $2 x 1 - x^{2} = 2 sin θ 1 - sin^{2} θ = 2 sin θ cos θ = sin 2 θ$ . So $sin^{- 1} (2 x 1 - x^{2}) = sin^{- 1} (sin 2 θ) = 2 θ = 2 sin^{- 1} x$ — valid here because the given range on $x$ keeps $2 θ$ inside $[- \frac{π}{2}, \frac{π}{2}]$ .

The same idea handles all four substitutions the book actually uses: $x = sin θ$ or $x = cos θ$ for expressions with $1 - x^{2}$ , $x = tan θ$ or $x = a tan θ$ for expressions with $1 + x^{2}$ or $a^{2} + x^{2}$ , and $x = sec θ$ for expressions with $x^{2} - 1$ .

Choosing the substitution from the radical. The pattern is mechanical once seen, because each choice is the one that makes the radical collapse by a Pythagorean identity:

Expression contains	Substitute	Radical becomes
$1 - x^{2}$	$x = sin θ$ (or $cos θ$ )	$cos θ$ (or $sin θ$ )
$1 + x^{2}$	$x = tan θ$	$sec θ$
$a^{2} + x^{2}$	$x = a tan θ$	$a sec θ$
$x^{2} - 1$	$x = sec θ$	$tan θ$

The identities the collapsed forms then need are the standard double-angle results, which is why the answers come out as multiples of $θ$ :

$sin 2 θ = 2 sin θ cos θ, cos 2 θ = 1 - 2 sin^{2} θ = 2 cos^{2} θ - 1, tan 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ}$

The range check is not optional. The final step $sin^{- 1} (sin 2 θ) = 2 θ$ is only valid while $2 θ$ stays inside $[- \frac{π}{2}, \frac{π}{2}]$ . This is exactly why the textbook attaches a condition such as $- \frac{1}{2} \leq x \leq \frac{1}{2}$ to the identity — that restriction on $x$ is what keeps $2 θ$ in range. Quoting the identity without its condition is only half the answer.

A worked half-angle case. Simplify $tan^{- 1} \frac{x}{1 + 1 - x ^{2}}$ . Put $x = sin θ$ , so $1 - x^{2} = cos θ$ and the expression becomes $tan^{- 1} \frac{sin θ}{1 + cos θ}$ . Using $sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}$ and $1 + cos θ = 2 cos^{2} \frac{θ}{2}$ , the fraction reduces to $tan \frac{θ}{2}$ , giving the answer $\frac{θ}{2} = \frac{1}{2} sin^{- 1} x$ .

Summary

Restricting a trig function's domain to its principal value branch makes it one-one and onto, so its inverse is well defined there.
$sin^{- 1} x$ is never the same as $(sin x)^{- 1} = 1/ sin x$ .
$sin (sin^{- 1} x) = x$ for $x \in [- 1, 1]$ ; $sin^{- 1} (sin x) = x$ only for $x$ already inside the principal branch $[- \frac{π}{2}, \frac{π}{2}]$ — outside it, first find the equivalent angle that does lie in the branch.
Every simplification substitutes $x$ as a trig ratio of a new angle (sin, cos, tan, or sec), reduces using a standard identity, and reads off the multiple of that angle — checking the resulting angle actually sits inside the target principal branch.
The current edition does not box a list of identities like sin⁻¹x+cos⁻¹x=π/2 — that is left-over content from older editions, not something this chapter states or requires.
Each principal branch is the shortest interval on which the function passes through all its values monotonically, which is why the sine and cosine branches differ.
$tan^{- 1}$ and $cot^{- 1}$ have open ranges; $cosec^{- 1}$ excludes $0$ and $sec^{- 1}$ excludes $\frac{π}{2}$ because those functions are undefined there.
$sin^{- 1} (- x) = - sin^{- 1} x$ and $tan^{- 1} (- x) = - tan^{- 1} x$ , but $cos^{- 1} (- x) = π - cos^{- 1} x$ — the odd-function rule does not apply to $cos^{- 1}$ or $cot^{- 1}$ .
Pick the substitution from the radical: $1 - x^{2} \to sin θ$ , $1 + x^{2} \to tan θ$ , $x^{2} - 1 \to sec θ$ .
An identity proved this way is only valid on the range of $x$ that keeps the resulting multiple of $θ$ inside the principal branch — quote the condition with the identity.
When solving equations in inverse trig functions, always substitute candidate solutions back into the original equation — the double/half-angle substitutions used to solve them can introduce extraneous roots that satisfy the transformed equation but not the original one.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Principal value branches

sin^{-1}: [-1,1] to [-pi/2,pi/2]; cos^{-1}: [-1,1] to [0,pi]; cosec^{-1}: R-(-1,1) to [-pi/2,pi/2]-{0}; sec^{-1}: R-(-1,1) to [0,pi]-{pi/2}; tan^{-1}: R to (-pi/2,pi/2); cot^{-1}: R to (0,pi)

Whenever a branch is not specified, it always means the principal value branch

sin^{-1}x is not (sin x)^{-1}

(sin x)^{-1} = 1/sin x, a completely different quantity from sin^{-1}x

The book states this explicitly as its own note — a very common notation confusion

Composition identities

sin(sin^{-1}x) = x for x in [-1,1]; sin^{-1}(sin x) = x only for x already in [-pi/2,pi/2]

Outside the principal branch, first find the angle inside the branch with the same sine value, using supplementary/periodic angle relations

Substitution technique

For expressions with sqrt(1-x^2), substitute x=sinθ or x=cosθ. For sqrt(1+x^2) or sqrt(a^2+x^2), substitute x=tanθ or x=a tanθ. For sqrt(x^2-1), substitute x=secθ

Reduces the composite expression to a single standard angle identity (double, half, or triple angle) — this is how every Exercise 2.2 question is actually solved

Triple angle formulas (used for proofs)

sin3θ = 3sinθ - 4sin^3θ; cos3θ = 4cos^3θ - 3cosθ; tan3θ = (3tanθ-tan^3θ)/(1-3tan^2θ)

These are the Class 11 identities this chapter's proof questions (like 3sin^{-1}x=sin^{-1}(3x-4x^3)) are built on

Half-angle identities (used for proofs)

(1-cos x)/(1+cos x) = tan^2(x/2); 1+sin x = (cos(x/2)+sin(x/2))^2; 1-sin x = (cos(x/2)-sin(x/2))^2

The exact identities behind Exercise 2.2 Q4 and Miscellaneous Q9's simplifications

Standard principal values

sin^{-1}(1/2)=pi/6; sin^{-1}(1/sqrt2)=pi/4; sin^{-1}(sqrt3/2)=pi/3; cos^{-1}(1/2)=pi/3; tan^{-1}(1)=pi/4; tan^{-1}(sqrt3)=pi/3

The handful of standard-angle values that most Exercise 2.1 questions reduce to once the right reference angle is identified

Graph as a reflection

The graph of y=f^{-1}(x) is the mirror image of y=f(x) in the line y=x

Applies to every inverse trig graph in this chapter — if (a,b) is on y=sin x, then (b,a) is on y=sin^{-1}x

Why each branch is chosen

The principal branch is the shortest interval on which the function runs monotonically through all of its values exactly once

Sine climbs from -1 to 1 on [-pi/2, pi/2]; cosine must instead fall from 1 to -1 on [0, pi], which is why the two branches differ

Open ranges for arctan and arccot

arctan has range (-pi/2, pi/2) and arccot has range (0, pi), both OPEN intervals

tan and cot run off to infinity at the endpoints and never attain a value there, so writing a closed interval is a marked error

Excluded points in the arccosec and arcsec ranges

arccosec excludes 0 from its range; arcsec excludes pi/2

Those are exactly the points where cosec and sec are undefined, because their reciprocals vanish there

Negative arguments, odd-function branches

arcsin(-x) = -arcsin(x) and arctan(-x) = -arctan(x)

Valid because these branches are symmetric about the origin

Negative arguments, the other branches

arccos(-x) = pi - arccos(x), and arccot(-x) = pi - arccot(x)

Their ranges sit inside [0, pi], so the odd-function rule does NOT apply — a common and costly slip

Choosing the substitution from the radical

sqrt(1-x^2) suggests x = sin(theta); sqrt(1+x^2) suggests x = tan(theta); sqrt(x^2-1) suggests x = sec(theta); sqrt(a^2+x^2) suggests x = a tan(theta)

Each choice is the one that collapses the radical by a Pythagorean identity

Validity condition on a proved identity

An identity such as arcsin(2x sqrt(1-x^2)) = 2 arcsin(x) holds only while the resulting multiple of theta stays inside the principal branch

That is why the textbook attaches a range on x — quoting the identity without its condition is only half the answer

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Assuming sin^{-1}(sin x) always equals x

✓ This only holds when x is already inside the principal branch [-pi/2,pi/2]. For x outside it, find the angle inside the branch with the same sine value first (e.g. sin(3pi/5)=sin(pi-3pi/5)=sin(2pi/5), and 2pi/5 is inside the branch).

WATCH OUT

✗ Reading sin^{-1}x as 1/sin x

✓ sin^{-1}x is the inverse function (an angle); 1/sin x is (sin x)^{-1}, the reciprocal (a ratio). They are unrelated quantities that happen to look similar in notation.

WATCH OUT

✗ Substituting x=sinθ (or cosθ, tanθ, secθ) without tracking which interval θ must lie in

✓ The interval restriction is what guarantees the final answer lands inside the correct principal branch — skipping it can silently produce a wrong-branch answer that looks algebraically correct.

WATCH OUT

✗ Accepting every algebraic root when solving an inverse-trig equation

✓ Squaring or using a double-angle identity during the solve can introduce extraneous roots. Always substitute each candidate back into the original equation before finalising the answer.

WATCH OUT

✗ Trying to quote a memorised list of 'properties' like sin^{-1}x+cos^{-1}x=pi/2 as if the book states them

✓ The current edition doesn't box such a list at all — every simplification is done via the substitution technique instead. Revise the technique, not a formula sheet that isn't actually there.

WATCH OUT

✗ Applying the odd-function rule to arccos or arccot

✓ arccos(-x) = pi - arccos(x), not -arccos(x). Only the branches symmetric about the origin, arcsin and arctan, negate through.

WATCH OUT

✗ Writing the range of arctan as a closed interval [-pi/2, pi/2]

✓ It is open, (-pi/2, pi/2), because tan is undefined at both endpoints. The same applies to arccot with (0, pi).

WATCH OUT

✗ Quoting a proved identity without the range of x on which it is valid

✓ Identities in this chapter are conditional. State the restriction on x that keeps the resulting angle inside the principal branch, or the result is only partially correct.

WATCH OUT

✗ Forgetting that arccosec excludes 0 and arcsec excludes pi/2 from their ranges

✓ Those points are removed because cosec and sec are undefined there. A final answer landing on an excluded value signals an error earlier in the working.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 2.1

Exercise 2.1

Finding principal values of inverse trig functions applied to standard ratios, and evaluating sums of principal values

Questions

Ex 2.2

Exercise 2.2

Proving identities and simplifying composite inverse-trig expressions using the substitution technique

Questions

Miscellaneous Ex

Miscellaneous Exercise

Mixed proofs combining inverse trig sums, simplification, and solving equations involving inverse trig functions

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Inverse Trigonometric Functions?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~8 marks in CBSE exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A trig function is invertible only after its domain is restricted to a principal value branch where it is one-one and onto
•sin^{-1}x is never the same as (sin x)^{-1} = 1/sin x
•sin(sin^{-1}x)=x always (for x in [-1,1]); sin^{-1}(sin x)=x only when x is already inside [-pi/2,pi/2]
•The four substitutions used throughout: x=sinθ or cosθ for sqrt(1-x^2), x=tanθ for sqrt(1+x^2), x=secθ for sqrt(x^2-1)
•Triple angle formulas (sin3θ, cos3θ, tan3θ) are what the book's 'prove that 3sin^{-1}x=...' style questions are built on
•The current edition has no boxed list of properties like sin^{-1}x+cos^{-1}x=pi/2 — every simplification goes through the substitution technique instead
•When solving an inverse-trig equation, always verify the final answer against the original equation — the solving process can introduce extraneous roots

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit I: 8 marks, shared with Relations and Functions

Question type	Marks each	Typical count	What it tests
Principal Values and Composition Identities	1-3	1	Finding principal values of standard ratios, evaluating sin^{-1}(sin x) and cos^{-1}(cos x) for x outside the principal branch
Simplification via Substitution	3-5	1	Substituting x=sinθ/cosθ/tanθ/secθ and reducing with a standard trig identity
Proofs and Solving Equations	4-6	1	Proving sums of inverse trig terms equal a target expression, solving equations and checking for extraneous roots

Prep strategy

Memorise the six domain/range pairs cold — a wrong branch silently produces a wrong-sign or wrong-quadrant answer that looks correct algebraically
For any sin^{-1}(sin x) or cos^{-1}(cos x) question where x looks unusual (like 7pi/6), first find the equivalent angle inside the principal branch before answering
For proof and solve questions, always substitute the final answer back into the original (not the transformed) equation to catch extraneous roots

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Robotics and inverse kinematics

Computing the joint angle needed to reach a target position requires inverse trig functions — given a ratio of arm-segment lengths, sin^{-1} or cos^{-1} recovers the actual angle a motor must turn to.

Navigation and bearing calculations

Given the north-south and east-west distance components between two points, tan^{-1} of their ratio recovers the compass bearing — exactly the same 'ratio to angle' operation this chapter formalises.

Signal processing phase recovery

Recovering the phase angle of a signal from its sine or cosine component uses inverse trig functions, with the same principal-branch care needed here to avoid a wrapped or ambiguous phase reading.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write out the domain and range table from memory before starting any problem set — most silent errors in this chapter come from an answer landing outside the correct principal branch

For 'find the value' questions on standard ratios, identify the reference angle first, then adjust its sign/quadrant to match the principal branch

For simplification questions, scan for sqrt(1-x^2), sqrt(1+x^2), or sqrt(x^2-1) patterns first — they signal exactly which substitution to use

For solve-the-equation questions, budget time to substitute every candidate root back into the original equation — this chapter's MCQs are specifically designed to catch students who stop at the algebra

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

The general (non-principal) solutions of inverse trig equations, involving all branches rather than just the principal one, extend this chapter's single-branch focus to the full periodic family of solutions

STRETCH

Inverse hyperbolic functions (sinh^{-1}, cosh^{-1}, tanh^{-1}) follow an entirely parallel construction to this chapter's inverse trig functions, using hyperbolic identities in place of ordinary trig identities

STRETCH

Complex inverse trig functions, where sin^{-1} and cos^{-1} are extended to complex arguments via logarithms, generalise the principal-branch idea beyond the real number line entirely

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainExtraneous roots in an inverse-trig equationSolve and verify against the original equation

Solve for $x$ : $sin^{- 1} (1 - x) - 2 sin^{- 1} x = \frac{π}{2}$ .

Stuck? Show the approach

Let $sin^{- 1} x = α$ , rewrite the equation in terms of $α$ , use $cos (2 α) = 1 - 2 sin^{2} α$ , solve the resulting polynomial in $x$ , then check every root against the original equation.

Show the full solution

With $x = sin α$ : $sin^{- 1} (1 - x) = \frac{π}{2} + 2 α \Rightarrow 1 - x = sin (\frac{π}{2} + 2 α) = cos 2 α = 1 - 2 x^{2}$ . So $- x = - 2 x^{2} \Rightarrow 2 x^{2} - x = 0 \Rightarrow x (2 x - 1) = 0 \Rightarrow x = 0$ or $x = \frac{1}{2}$ . Checking $x = \frac{1}{2}$ in the original equation: $sin^{- 1} (\frac{1}{2}) - 2 sin^{- 1} (\frac{1}{2}) = \frac{π}{6} - \frac{π}{3} = - \frac{π}{6} \neq = \frac{π}{2}$ — extraneous. Checking $x = 0$ : $sin^{- 1} (1) - 0 = \frac{π}{2}$ — valid.

Answer: x = 0

The trap

Stopping at the algebraic roots $x = 0, \frac{1}{2}$ without substituting back into the original equation — $x = \frac{1}{2}$ satisfies the squared/transformed version but not the actual equation, exactly the kind of extraneous root this equation is designed to test for.

JEE MainSum of two inverse-sine terms via the addition techniqueRight-triangle ratio extraction plus the sine addition formula

Prove that $sin^{- 1} \frac{8}{17} + sin^{- 1} \frac{3}{5} = tan^{- 1} \frac{77}{36}$ .

Stuck? Show the approach

Convert each inverse sine to its full set of ratios (sin, cos, tan) via a right triangle, then combine using the tangent addition formula, checking the resulting angle sum stays inside the principal branch.

Show the full solution

Let $α = sin^{- 1} \frac{8}{17}$ : $sin α = \frac{8}{17}$ , $cos α = \frac{15}{17}$ , $tan α = \frac{8}{15}$ . Let $β = sin^{- 1} \frac{3}{5}$ : $tan β = \frac{3}{4}$ . $tan (α + β) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}} = \frac{77/60}{3/5} = \frac{77}{36}$ . Since $α \approx 2 8^{\circ}$ and $β \approx 3 7^{\circ}$ , their sum is under $9 0^{\circ}$ , inside the principal branch of $tan^{- 1}$ .

Answer: Identity proved: both sides equal alpha+beta

The trap

Forgetting to check that $α + β$ actually lands inside $(- \frac{π}{2}, \frac{π}{2})$ before equating to $tan^{- 1}$ of the combined ratio — for larger inputs this same technique would need a $π$ correction.

JEE MainSimplification via the secant substitutionSubstitution reducing a composite expression to a single inverse trig term

Simplify $sec^{- 1} (\frac{1}{2 x ^{2} - 1})$ in terms of $cos^{- 1} x$ , for $0 < x < \frac{1}{2}$ .

Stuck? Show the approach

Substitute $x = cos θ$ and recognise $2 x^{2} - 1$ as the cosine double-angle formula.

Show the full solution

Let $x = cos θ$ . Then $2 x^{2} - 1 = 2 cos^{2} θ - 1 = cos 2 θ$ , so $\frac{1}{2 x ^{2} - 1} = \frac{1}{cos 2 θ} = sec 2 θ$ . So the expression becomes $sec^{- 1} (sec 2 θ) = 2 θ$ , valid since the given range on $x$ keeps $2 θ$ inside $[0, π] - {π /2}$ .

Answer: 2cos^{-1}x

The trap

Missing that $2 x^{2} - 1$ is exactly the double-angle cosine formula in disguise — without spotting that, the expression looks unsimplifiable.

JEE MainSolving a tan-inverse-sum equationAddition formula equated to a target angle

Solve for $x > 0$ : $tan^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} tan^{- 1} x$ .

Stuck? Show the approach

Recognise the left side as the standard form $tan^{- 1} 1 - tan^{- 1} x = \frac{π}{4} - tan^{- 1} x$ , then solve the resulting linear equation in $tan^{- 1} x$ .

Show the full solution

$\frac{1 - x}{1 + x} = \frac{tan \frac{π}{4} - tan α}{1 + tan \frac{π}{4} tan α} = tan (\frac{π}{4} - α)$ where $x = tan α$ . So the equation becomes $\frac{π}{4} - tan^{- 1} x = \frac{1}{2} tan^{- 1} x$ , giving $\frac{π}{4} = \frac{3}{2} tan^{- 1} x$ , so $tan^{- 1} x = \frac{π}{6}$ , giving $x = tan \frac{π}{6} = \frac{1}{3}$ .

Answer: x = 1/sqrt3

The trap

Not recognising $\frac{1 - x}{1 + x}$ as the tangent-subtraction pattern with $1 = tan \frac{π}{4}$ — attempting to cross-multiply and solve directly leads to a much messier equation.

JEE AdvancedGeneral solution combining two inverse-tan constraintsMulti-step equation with a domain restriction check

If $tan^{- 1} x + tan^{- 1} y + tan^{- 1} z = π$ and $x, y, z$ are all positive, show that $x + y + z = x y z$ .

Stuck? Show the approach

Rearrange to $tan^{- 1} x + tan^{- 1} y = π - tan^{- 1} z$ , take the tangent of both sides using $tan (π - θ) = - tan θ$ , and apply the addition formula.

Show the full solution

$tan (tan^{- 1} x + tan^{- 1} y) = tan (π - tan^{- 1} z) = - tan (tan^{- 1} z) = - z$ . The left side, by the addition formula, is $\frac{x + y}{1 - x y}$ (valid since $x, y > 0$ makes each individual angle positive, and their sum together with the third angle must total $π$ , so $1 - x y \neq = 0$ ). So $\frac{x + y}{1 - x y} = - z \Rightarrow x + y = - z (1 - x y) = - z + x y z \Rightarrow x + y + z = x y z$ .

Answer: Identity proved: x+y+z=xyz

The trap

Applying the addition formula $tan^{- 1} x + tan^{- 1} y = tan^{- 1} (\frac{x + y}{1 - x y})$ directly without first isolating one term on the other side — taking the tangent of the rearranged equation is what avoids the branch ambiguity of adding three inverse-tan terms directly.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 BoardMedium

JEE MainHigh

JEE AdvancedMedium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The current book doesn't box or use these identities directly — every exercise question is solved through the sinθ/cosθ/tanθ/secθ substitution technique shown in the worked examples instead. Focus your revision there.

Because sin^{-1} only undoes sin exactly when the angle you started with was already inside the principal branch [-pi/2,pi/2]. For any x outside it, first find the angle inside the branch that has the same sine value (using symmetry like sin(pi-x)=sin x), then apply sin^{-1}(sin of that angle)=that angle.

Look at what's under the square root, if there is one: sqrt(1-x^2) suggests x=sinθ or cosθ; sqrt(1+x^2) or sqrt(a^2+x^2) suggests x=tanθ or a tanθ; sqrt(x^2-1) suggests x=secθ. If there's no square root, look for a pattern matching a double, half, or triple angle formula directly.

Squaring or using identities like cos2θ=1-2sin²θ during the solving process can introduce extraneous roots that satisfy the transformed equation but not the original one. Always plug every candidate root back into the original equation before finalising an answer.

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