CUET UG Mathematics Booster Test 1 - Inverse Trigonometric Functions -

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CUET UG Mathematics Booster Test 1 - Inverse Trigonometric Functions

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Mathematics Booster Test 1 - Inverse Trigonometric Functions

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QUESTION 1 OF 20

If f(x) is an invertible function with domain X and range Y, what are the domain and range of its inverse g?

Answer Review

1 If f(x) is an invertible function with domain X and range Y, what are the domain and range of its inverse g?

A Domain X, Range Y

B Domain Y, Range X

C Domain Y, Range Y

D Domain X, Range X

💡 Short Explanation

Inverse swaps inputs and outputs. Domain of inverse equals original range. Range of inverse equals original domain.

✅ Correct Answer Explanation

For an invertible function f: X → Y, the inverse function reverses mappings. Hence the domain of g=f⁻¹ becomes Y and the range becomes X. Option B correctly states this interchange. Options A, C, and D ignore the reversal property fundamental to inverse functions.

❓ Why Other Options Are Incorrect

Option A → Keeps domain and range unchanged, which is incorrect for inverses.
Option C → Both domain and range cannot simultaneously equal Y.
Option D → Both domain and range cannot simultaneously equal X.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall that inverse functions reverse the original mapping direction.

Final Logic:

Inverse functions interchange domain and range.

🧠 Memory Trick

"Inverse flips domain and range."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Inverse Functions

2 Match the inverse trigonometric function to its principal value branch range:

List 1	List 2
1. cos⁻¹	a. (-π/2, π/2)
2. cosec⁻¹	b. (0, π)
3. tan⁻¹	c. [0, π]
4. cot⁻¹	d. [-π/2, π/2] - {0}

A 1-b, 2-c, 3-a, 4-d

B 1-a, 2-d, 3-b, 4-c

C 1-c, 2-d, 3-a, 4-b

D 1-d, 2-a, 3-c, 4-b

💡 Short Explanation

cos⁻¹ range is [0,π]. tan⁻¹ range is (-π/2,π/2). cot⁻¹ range is (0,π).

✅ Correct Answer Explanation

The standard principal ranges are: cos⁻¹x → [0,π], cosec⁻¹x → [-π/2,π/2]−{0}, tan⁻¹x → (-π/2,π/2), cot⁻¹x → (0,π). Thus the correct matching is 1-c, 2-d, 3-a, 4-b, making option C correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly assigns cosec⁻¹ and cot⁻¹ ranges.
Option B → Misplaces cos⁻¹ and tan⁻¹ ranges.
Option D → cos⁻¹ does not have range excluding zero.

🎯 Strategy

Used: Option Grouping

Application:

Match each inverse trigonometric function with its standard NCERT principal branch.

Final Logic:

Correct range identification gives the exact matching.

🧠 Memory Trick

"Cos: [0,π], Tan: open ±π/2."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Value Branches

3 For the tangent function, which intervals represent restricted domains where the function becomes bijective?
I. (-π/2, π/2)
II. (π/2, 3π/2)
III. (-3π/2, -π/2)

A I only

B I and II only

C II and III only

D I, II, and III

💡 Short Explanation

Tangent is monotonic between asymptotes. Each interval avoids discontinuities. Hence tangent becomes bijective there.

✅ Correct Answer Explanation

The tangent function is continuous and strictly increasing between consecutive vertical asymptotes. Therefore intervals like (-π/2,π/2), (π/2,3π/2), and (-3π/2,-π/2) all make tangent bijective. Hence statements I, II, and III are correct, making option D the correct answer.

❓ Why Other Options Are Incorrect

Option A → Ignores other valid bijective intervals.
Option B → Excludes interval III unnecessarily.
Option C → Omits the standard principal branch interval.

🎯 Strategy

Used: Elimination

Application:

Check whether tangent remains one-one and onto in each interval.

Final Logic:

Every interval between consecutive asymptotes gives bijection.

🧠 Memory Trick

"Between asymptotes, tan behaves perfectly."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Restricted Domains of Tangent

4 Which of the following statements is incorrect regarding the function y = sec⁻¹ x?

A Its domain is the closed interval [-1, 1].

B Its principal value branch is [0, π] - {π/2}.

C It represents the angle whose secant is x.

D It is undefined for values strictly between -1 and 1.

💡 Short Explanation

sec⁻¹x exists for |x|≥1. Values between -1 and 1 are invalid. Hence option A is false.

✅ Correct Answer Explanation

The inverse secant function is defined only for x≤-1 or x≥1. Therefore its domain is (-∞,-1]∪[1,∞), not [-1,1]. Option B correctly gives the principal branch. Option C correctly describes inverse secant. Option D is also true because secant never lies strictly between -1 and 1.

❓ Why Other Options Are Incorrect

Option B → Correct principal branch for sec⁻¹x.
Option C → Correct interpretation of inverse secant.
Option D → True because secant values satisfy |sec x|≥1.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether the stated interval fully satisfies secant output restrictions.

Final Logic:

sec⁻¹x cannot accept values inside (-1,1).

🧠 Memory Trick

"Sec inverse starts outside ±1."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domain of Secant Inverse

5 In a calculus problem involving numerical substitution, the expression 3sin⁻¹(x) is modeled. According to the properties of inverse trigonometric functions, 3sin⁻¹(x) = sin⁻¹(3x - 4x³). At x = 1/2, calculate the numerical value of 3*sin⁻¹(x) in radians.

A π/6

B π/2

C π/3

D π

💡 Short Explanation

sin⁻¹(1/2)=π/6 Multiply by 3. Result becomes π/2.

✅ Correct Answer Explanation

At x=1/2, sin⁻¹(1/2)=π/6 because sine of π/6 equals 1/2. Therefore: 3sin⁻¹(1/2)=3×π/6=π/2. Thus option B is correct. Option A gives only the inverse sine value before multiplication. Options C and D produce incorrect multiples.

❓ Why Other Options Are Incorrect

Option A → Represents only sin⁻¹(1/2), not three times it.
Option C → Incorrect multiplication of π/6.
Option D → Exceeds the required calculated value.

🎯 Strategy

Used: Substitution

Application:

Insert x=1/2 directly into the expression and simplify.

Final Logic:

Three times π/6 equals π/2.

🧠 Memory Trick

"Half gives π/6; triple gives π/2."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values

6 An engineering plot requires finding the principal value of cot⁻¹(-1/√3). Which region of the graph will the angle lie in?

A First quadrant (0 to π/2)

B Fourth quadrant (-π/2 to 0)

C Second quadrant (π/2 to π)

D Third quadrant (-π to -π/2)

💡 Short Explanation

cot⁻¹ range is (0,π). Negative cotangent occurs in second quadrant. Hence angle lies in quadrant II.

✅ Correct Answer Explanation

The principal value range of cot⁻¹x is (0,π). Cotangent is negative in the second quadrant. Since cot(2π/3)=-1/√3, the required angle lies in the second quadrant. Therefore option C is correct. Other quadrants either produce positive cotangent or lie outside the principal range.

❓ Why Other Options Are Incorrect

Option A → Cotangent is positive in first quadrant.
Option B → Fourth quadrant is not within cot⁻¹ principal range.
Option D → Third quadrant also lies outside the principal branch interval.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use sign conventions of cotangent together with principal branch intervals.

Final Logic:

Negative cotangent within (0,π) belongs to quadrant II.

🧠 Memory Trick

"cot⁻¹ lives in (0,π)."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of Cotangent Inverse

7 Consider the dataset of the principal values:
a = sin⁻¹(1/2),
b = sin⁻¹(-1/2),
c = cos⁻¹(1/2).
Calculate the arithmetic mean (average) of these values.

A π/6

B π/4

C π/3

D π/9

💡 Short Explanation

Values are π/6, -π/6, π/3. Sum equals π/3. Divide by 3 to get π/9.

✅ Correct Answer Explanation

The principal values are: sin⁻¹(1/2)=π/6, sin⁻¹(-1/2)=-π/6, cos⁻¹(1/2)=π/3. Their sum is π/3. Dividing by 3 gives π/9. Therefore option D is correct. The other options result from incorrect addition or averaging of the standard angles.

❓ Why Other Options Are Incorrect

Option A → Ignores cancellation between positive and negative terms.
Option B → Incorrect averaging after summation.
Option C → Represents only one of the individual values.

🎯 Strategy

Used: Substitution

Application:

Replace inverse trigonometric expressions with standard principal values.

Final Logic:

Average equals (π/3)/3 = π/9.

🧠 Memory Trick

"π/6 and -π/6 cancel."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Standard Principal Values

8 From the set of values {-1, -1/2, 0, 1/2, 1}, an input x is chosen at random. What is the theoretical probability that cos⁻¹(x) results in an angle greater than or equal to π/2?

A 3/5

B 2/5

C 4/5

D 1/5

💡 Short Explanation

cos⁻¹x ≥ π/2 when x≤0. Valid values: -1, -1/2, 0. Probability = 3/5.

✅ Correct Answer Explanation

The cosine inverse function decreases from 0 to π as x increases from 1 to -1. Therefore cos⁻¹(x)≥π/2 whenever x≤0. From the given set, the valid values are -1, -1/2, and 0. Hence probability = 3/5, making option A correct.

❓ Why Other Options Are Incorrect

Option B → Counts only two valid values instead of three.
Option C → Includes one positive value incorrectly.
Option D → Greatly undercounts satisfying outcomes.

🎯 Strategy

Used: Elimination

Application:

Identify which x-values produce angles at least π/2.

Final Logic:

Three out of five numbers satisfy x≤0.

🧠 Memory Trick

"Cos inverse ≥90° means x nonpositive."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Range of Cosine Inverse

9 Vector v is defined as v = tan⁻¹(1) i + tan⁻¹(-1) j. What is the numerical sum of its components?

A π/2

B 0

C π

D -π/2

💡 Short Explanation

tan⁻¹(1)=π/4 tan⁻¹(-1)=-π/4 Sum equals zero.

✅ Correct Answer Explanation

The principal values are tan⁻¹(1)=π/4 and tan⁻¹(-1)=-π/4. Adding them gives: π/4 + (-π/4)=0. Hence option B is correct. Options A, C, and D result from incorrect handling of signs or principal values.

❓ Why Other Options Are Incorrect

Option A → Represents only positive contribution doubled.
Option C → Far exceeds the actual sum.
Option D → Incorrectly ignores the positive component.

🎯 Strategy

Used: Substitution

Application:

Replace inverse tangent terms using standard principal values.

Final Logic:

Equal positive and negative values cancel completely.

🧠 Memory Trick

"π/4 and -π/4 balance."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Standard Tangent Inverse Values

10 The area of a triangle is given by A = (1/2) * base * height. If base = cot⁻¹(1) and height = 4/π, what is the area?

A 1

B π/2

C 1/2

D π/4

💡 Short Explanation

cot⁻¹(1)=π/4 Use triangle area formula. Result simplifies to 1/2.

✅ Correct Answer Explanation

The principal value cot⁻¹(1)=π/4. Therefore: Area = (1/2)×(π/4)×(4/π). The π and 4 cancel, leaving: Area = 1/2. Thus option C is correct. The remaining options arise from incomplete simplification or formula misuse.

❓ Why Other Options Are Incorrect

Option A → Ignores the factor 1/2 in the area formula.
Option B → Incorrect simplification involving π terms.
Option D → Represents only the base value.

🎯 Strategy

Used: Substitution

Application:

Substitute cot⁻¹(1)=π/4 into the area formula directly.

Final Logic:

Cancellation simplifies the area exactly to 1/2.

🧠 Memory Trick

"π cancels completely."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of Cotangent Inverse

11 An integral evaluates to the boundary expression y = sec⁻¹(2/√3). What is the exact value of y?

A π/3

B π/4

C π/2

D π/6

💡 Short Explanation

sec y = 2/√3 Therefore cos y = √3/2 Principal angle is π/6

✅ Correct Answer Explanation

Given sec⁻¹(2/√3)=y, we write sec y=2/√3. Taking reciprocal gives cos y=√3/2. The principal angle satisfying this is π/6. Hence option D is correct. Option A gives cosine 1/2, option B gives √2/2, and option C gives cosine zero, where secant is undefined.

❓ Why Other Options Are Incorrect

Option A → sec(π/3)=2, not 2/√3.
Option B → sec(π/4)=√2, which does not match.
Option C → secant is undefined at π/2 because cos(π/2)=0.

🎯 Strategy

Used: Substitution

Application:

Convert secant inverse into cosine form using reciprocal identities.

Final Logic:

cos y=√3/2 corresponds to y=π/6.

🧠 Memory Trick

"√3/2 means 30°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of Secant Inverse

12 Assertion (A): The principal value of cosec⁻¹(2) is π/6.
Reason (R): cosec(π/6) = 2 and π/6 lies in the interval [-π/2, π/2] - {0}.

A Both true, R explains A

B A true, R false

C A false, R true

D Both false

💡 Short Explanation

cosec(π/6)=2 π/6 belongs to principal branch Hence assertion and reason are true

✅ Correct Answer Explanation

The inverse cosecant function uses the principal branch [-π/2,π/2]−{0}. Since cosec(π/6)=2 and π/6 lies in the principal interval, cosec⁻¹(2)=π/6. Therefore both Assertion and Reason are true, and the Reason correctly explains the Assertion. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Reason is mathematically correct.
Option C → Assertion is true because cosec(π/6)=2.
Option D → Both statements are valid according to principal value definitions.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Verify both the trigonometric identity and principal branch condition together.

Final Logic:

The reason directly justifies the assertion.

🧠 Memory Trick

"cosec 30° = 2."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of Cosecant Inverse

13 Arrange the following expressions in ascending order of their analytical values:
1. sin⁻¹(sin 3π/5)
2. cos⁻¹(cos 13π/6)
3. tan⁻¹(tan π/4)
4. sin⁻¹(-1/2)

A 1, 2, 3, 4

B 2, 3, 4, 1

C 4, 2, 3, 1

D 4, 1, 3, 2

💡 Short Explanation

sin⁻¹(-1/2)=-π/6 cos⁻¹(cos13π/6)=π/6 tan⁻¹(tanπ/4)=π/4 and sin⁻¹(sin3π/5)=2π/5

✅ Correct Answer Explanation

Evaluate each expression: 1 → sin⁻¹(sin3π/5)=2π/5 2 → cos⁻¹(cos13π/6)=π/6 3 → tan⁻¹(tanπ/4)=π/4 4 → sin⁻¹(-1/2)=-π/6 Ascending order: -π/6 < π/6 < π/4 < 2π/5. Thus the order is 4,2,3,1, making option C correct.

❓ Why Other Options Are Incorrect

Option A → Places the largest value first incorrectly.
Option B → Does not begin with the negative value.
Option D → Places 2π/5 before π/4 incorrectly.

🎯 Strategy

Used: Substitution

Application:

Reduce each inverse trigonometric expression to its principal value.

Final Logic:

Numerical comparison gives order 4,2,3,1.

🧠 Memory Trick

"Negative first, then small positives."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Value Evaluation

14 To find the principal value branch of sec⁻¹, the domain of the secant function is restricted to:

A [0, π]

B [-π/2, π/2]

C [-π/2, π/2] - {0}

D [0, π] - {π/2}

💡 Short Explanation

sec x undefined at π/2 Principal branch avoids repetition Hence interval excludes π/2

✅ Correct Answer Explanation

The secant function becomes bijective on the interval [0,π] excluding π/2, where secant is undefined. This restriction defines the principal branch for sec⁻¹x. Option A incorrectly includes π/2. Options B and C correspond to tangent or cosecant-related intervals rather than secant inverse.

❓ Why Other Options Are Incorrect

Option A → Includes π/2 where secant is undefined.
Option B → Not the standard branch for secant inverse.
Option C → Corresponds more closely to cosecant inverse ranges.

🎯 Strategy

Used: Elimination

Application:

Remove intervals containing undefined secant values or repeated outputs.

Final Logic:

Only [0,π]−{π/2} gives bijection for secant.

🧠 Memory Trick

"Sec inverse skips π/2."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Restricted Domain of Secant

15 If the natural range of the cosine function is [-1, 1], what is the principal value of cos⁻¹(-1/2)?

A 2π/3

B π/3

C -π/3

D 5π/6

💡 Short Explanation

cos(2π/3)=-1/2 Principal range is [0,π] Hence principal value is 2π/3

✅ Correct Answer Explanation

The inverse cosine function has principal range [0,π]. Since cos(2π/3)=-1/2, the principal value of cos⁻¹(-1/2) is 2π/3. Option B gives cosine +1/2. Option C lies outside the principal range. Option D gives cosine -√3/2, not -1/2.

❓ Why Other Options Are Incorrect

Option B → cos(π/3)=1/2, not negative.
Option C → Negative angle not allowed in cosine inverse principal branch.
Option D → cos(5π/6)=-√3/2, not -1/2.

🎯 Strategy

Used: Substitution

Application:

Check standard cosine values within the principal interval [0,π].

Final Logic:

Only 2π/3 gives cosine equal to -1/2.

🧠 Memory Trick

"Cos inverse stays positive."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of Cosine Inverse

16 The range of the inverse function y = tan⁻¹ x is:

A [-π/2, π/2]

B (-π/2, π/2)

C (0, π)

D [0, π]

💡 Short Explanation

tan x undefined at ±π/2 Endpoints excluded from range Principal branch is open interval

✅ Correct Answer Explanation

The inverse tangent function has principal value range (-π/2,π/2). The endpoints are excluded because tangent is undefined there. Hence option B is correct. Option A incorrectly includes undefined endpoints. Options C and D correspond to cotangent-related intervals rather than tangent inverse.

❓ Why Other Options Are Incorrect

Option A → Includes ±π/2 where tangent is undefined.
Option C → Represents cot⁻¹ principal range.
Option D → Incorrect closed interval unrelated to tan⁻¹.

🎯 Strategy

Used: Elimination

Application:

Exclude angles where tangent function becomes undefined.

Final Logic:

tan⁻¹x always lies strictly between ±π/2.

🧠 Memory Trick

"Tan inverse never touches ±90°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Branch of Tangent Inverse

17 A function f: R → R defined by f(x) = sin x is not invertible because:

A It is one-one but not onto

B It is onto but not one-one

C It is neither one-one nor onto

D It is undefined at 0

💡 Short Explanation

Sine repeats values periodically. Range limited to [-1,1]. Hence neither injective nor surjective on R→R.

✅ Correct Answer Explanation

The sine function repeats values, so it is not one-one. Its range is only [-1,1], not all real numbers, so it is not onto for codomain R. Therefore f:R→R is neither injective nor surjective. Hence option C is correct. Option D is false because sine is defined everywhere.

❓ Why Other Options Are Incorrect

Option A → Sine is not one-one because of periodicity.
Option B → Sine is not onto when codomain is R.
Option D → sin0=0, so the function is perfectly defined at zero.

🎯 Strategy

Used: Elimination

Application:

Check both injective and surjective conditions independently.

Final Logic:

Failure of both conditions prevents invertibility.

🧠 Memory Trick

"Sine repeats and stays bounded."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Non-Invertibility of Sine

18 Given tan⁻¹(√3) = y, what is the analytical value of y?

A π/6

B π/4

C π/2

D π/3

💡 Short Explanation

tan(π/3)=√3 π/3 lies in principal range Hence inverse value is π/3

✅ Correct Answer Explanation

The principal range of tan⁻¹x is (-π/2,π/2). Since tan(π/3)=√3 and π/3 belongs to the principal interval, tan⁻¹(√3)=π/3. Therefore option D is correct. The remaining options produce tangent values 1/√3, 1, and undefined respectively.

❓ Why Other Options Are Incorrect

Option A → tan(π/6)=1/√3, not √3.
Option B → tan(π/4)=1.
Option C → tangent is undefined at π/2.

🎯 Strategy

Used: Substitution

Application:

Recall standard tangent values from trigonometric tables.

Final Logic:

Only π/3 gives tangent equal to √3.

🧠 Memory Trick

"√3 belongs to 60°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Standard Tangent Inverse Values

📄 Passage

We know that the domain of the tangent function is the set {x : x ∈ R and x ≠ (2n + 1)π/2, n ∈ Z} and the range is R. It means that tan function is not defined for odd multiples of π/2. If we restrict the domain of tangent function to (-π/2, π/2), then it is one-one and onto with its range as R. Thus tan⁻¹ can be defined as a function whose domain is R and range could be any of the intervals (-3π/2, -π/2), (-π/2, π/2), (π/2, 3π/2), etc.

From the passage, the branch with range (-π/2, π/2) is known as what for the function tan⁻¹?

A Principal value branch

B Restricted domain branch

C Natural range

D Asymptotic branch

💡 Short Explanation

tan⁻¹ uses a standard branch. Principal range is (-π/2,π/2). This branch defines uniqueness.

✅ Correct Answer Explanation

The inverse tangent function is conventionally defined using the range (-π/2,π/2), called the principal value branch. This interval ensures tangent remains one-one and onto. Therefore option A is correct. The other options are not standard NCERT terminology for inverse tangent definitions.

❓ Why Other Options Are Incorrect

Option B → Domain restriction applies to tangent, not the inverse range naming.
Option C → Natural range refers to tangent outputs, not inverse branches.
Option D → Asymptotic branch is not a standard mathematical term here.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the exact terminology associated with inverse trigonometric branches.

Final Logic:

(-π/2,π/2) is the standard principal branch for tan⁻¹.

🧠 Memory Trick

"Tan inverse lives between ±90°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Value Branch

📄 Passage

According to the passage, the domain of tan⁻¹ function is:

A (-π/2, π/2)

B R (Set of all real numbers)

C [-1, 1]

D R - [-1, 1]

💡 Short Explanation

Tangent range is all real numbers. Inverse domain equals original range. Hence tan⁻¹ domain is R.

✅ Correct Answer Explanation

The tangent function has range R on its restricted principal interval. Since the domain of an inverse equals the range of the original function, tan⁻¹x has domain R. Therefore option B is correct. Option A is the principal range of tan⁻¹, not its domain.

❓ Why Other Options Are Incorrect

Option A → Represents range of tan⁻¹, not its domain.
Option C → Domain restriction [-1,1] applies to sine and cosine inverses.
Option D → tan⁻¹ accepts every real number without exclusion.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate inverse domain directly to tangent function range.

Final Logic:

Since tan x covers all real numbers, tan⁻¹ accepts all real numbers.

🧠 Memory Trick

"Tan inverse takes every real input."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domain and Range of Tangent Inverse

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