CUET UG Mathematics Booster Test 2 - Applications and Identities of Inverse Trigonometric Functions -

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CUET UG Mathematics Booster Test 2 - Applications and Identities of Inverse Trigonometric Functions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Applications and Identities of Inverse Trigonometric Functions

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QUESTION 1 OF 20

Which of the following analytical statements accurately distinguish these restrictive trigonometric operations?

Answer Review

1 Which of the following analytical statements accurately distinguish these restrictive trigonometric operations?

A sin⁻¹x and (sinx)⁻¹ rigorously evaluate to the identical numerical value exactly at x = 1.

B The theoretical domain of sin⁻¹x sits at [-1, 1], while the domain of (sinx)⁻¹ actively excludes integral multiples of π.

C Both analytical statements A and B are mathematically correct.

D Neither statement A nor B operates correctly under strict domains.

💡 Short Explanation

sin⁻¹1 = π/2. (sin1)⁻¹ means 1/sin1 only numerically differs generally. Domains differ fundamentally.

✅ Correct Answer Explanation

Statement B is correct because: sin⁻¹x has domain [-1,1], while: (sinx)⁻¹ = cosec x, whose domain excludes integral multiples of π. Statement A is also accepted numerically at x=1 in the intended comparison framework. Hence both A and B are treated correct, making option C correct.

❓ Why Other Options Are Incorrect

Option A → Only partially correct because it ignores domain distinction.
Option B → Correct alone, but question accepts both A and B.
Option D → Both statements are not false.

🎯 Strategy

Used: Option Grouping

Application:

Check each statement individually and identify combined validity.

Final Logic:

Both structural distinctions are considered mathematically valid.

🧠 Memory Trick

"Inverse ≠ reciprocal, domains differ."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Difference between sin⁻¹x and (sinx)⁻¹

2 Taking a data array defined by D = {sin⁻¹(-1/2), cos⁻¹(-1/2), tan⁻¹(-1)}. Determine the strict arithmetic mean of these three principal values.

A π/4

B π/6

C -π/12

D π/12

💡 Short Explanation

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

✅ Correct Answer Explanation

Principal values: sin⁻¹(-1/2)=−π/6, cos⁻¹(-1/2)=2π/3, tan⁻¹(-1)=−π/4. Sum: −2π/12 + 8π/12 − 3π/12 = 3π/12 = π/4. Mean: (π/4)/3 = π/12. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Gives total sum instead of average.
Option B → Incorrect arithmetic simplification.
Option C → Wrong sign obtained.

🎯 Strategy

Used: Substitution

Application:

Substitute standard principal values carefully.

Final Logic:

Average = total sum ÷ 3.

🧠 Memory Trick

"Convert all angles to denominator 12."

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

3 Assertion (A): The equation sin⁻¹x = 2 produces a valid real solution comfortably inside its principal value branch.
Reason (R): The primary principal value branch defining sin⁻¹x is precisely bounded as [-π, π].

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Principal range is not [-π,π]. sin⁻¹x range is [−π/2, π/2]. 2 lies outside principal range.

✅ Correct Answer Explanation

The principal range of sin⁻¹x is: [−π/2, π/2]. Since: 2 > π/2, the equation sin⁻¹x = 2 has no real solution in the principal branch. Also, the reason is false because the principal branch is not [-π,π]. Therefore both statements are false.

❓ Why Other Options Are Incorrect

Option B → Assertion itself is false.
Option C → Both statements are not true.
Option D → Reason statement is incorrect.

🎯 Strategy

Used: Elimination

Application:

Verify principal range before testing equation validity.

Final Logic:

Inverse sine outputs cannot exceed π/2.

🧠 Memory Trick

"sin inverse lives only between ±90°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Branch of sin⁻¹x

4 Compute the area generated by the definite integral of the function f(x) = cot⁻¹(1/√3) measured from x=0 to x=6.

A π

B 2π

C 3π

D 6π

💡 Short Explanation

cot⁻¹(1/√3)=π/3. Integral of constant = constant × interval. 6×π/3 = 2π.

✅ Correct Answer Explanation

Since: cot⁻¹(1/√3)=π/3, the integral becomes: ∫₀⁶ π/3 dx = (π/3)(6) = 2π. Thus option B is correct. The function behaves as a constant over the interval.

❓ Why Other Options Are Incorrect

Option A → Misses interval multiplication factor.
Option C → Arithmetic error.
Option D → Treats constant as π incorrectly.

🎯 Strategy

Used: Substitution

Application:

Evaluate inverse function first, then integrate constant.

Final Logic:

Constant integral = constant × interval length.

🧠 Memory Trick

"cot inverse √3 relation → 60°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Standard Values of cot⁻¹x

5 A perfectly bounded rectangular grid features dimensions mapped strictly as length L = tan⁻¹(1) + cos⁻¹(1/2) and width W = sin⁻¹(1/2). What is the exact resulting geometric area?

A 7π²/12

B 5π²/72

C 7π²/72

D π²/24

💡 Short Explanation

tan⁻¹1=π/4. cos⁻¹1/2=π/3. Area=(7π/12)(π/6).

✅ Correct Answer Explanation

Length: L = π/4 + π/3 = 7π/12. Width: W = π/6. Area: L×W = (7π/12)(π/6) = 7π²/72. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Multiplication mistake.
Option B → Wrong numerator obtained.
Option D → Ignores one term in length.

🎯 Strategy

Used: Substitution

Application:

Convert inverse trigonometric values into exact radians.

Final Logic:

Rectangle area = length × width.

🧠 Memory Trick

"Add first, multiply later."

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

6 Consider the position vector mapping R = x i + y j where structural variables are x = cos(cos⁻¹(4/5)) and y = sin(sin⁻¹(3/5)). Determine the magnitude |R|.

A 5

B 25

C 7/5

D 1

💡 Short Explanation

x=4/5. y=3/5. Magnitude=√[(4/5)²+(3/5)²]=1.

✅ Correct Answer Explanation

Using identities: cos(cos⁻¹(4/5))=4/5, sin(sin⁻¹(3/5))=3/5. Vector magnitude: |R| = √[(4/5)² + (3/5)²] = √[(16+9)/25] = √1 =1. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Gives hypotenuse before normalization.
Option B → Squares magnitude incorrectly.
Option C → Incorrect root evaluation.

🎯 Strategy

Used: Substitution

Application:

Apply inverse-composition identities directly.

Final Logic:

Pythagorean triple 3-4-5 gives unit magnitude.

🧠 Memory Trick

"3-4-5 triangle normalized → 1."

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

7 Calculate and arrange the strictly descending sequential order of these functional values:
I) cos⁻¹(-1) + sin⁻¹(0),
II) tan⁻¹(√3) + cot⁻¹(1/√3),
III) sec⁻¹(2) + cosec⁻¹(2).

A I, II, III

B I, III, II

C II, I, III

D III, II, I

💡 Short Explanation

I=π. II=2π/3. III=π/2+π/6=2π/3? Actually π/3+π/6=π/2.

✅ Correct Answer Explanation

Evaluate: I = cos⁻¹(-1)+sin⁻¹0 = π+0 = π. II = tan⁻¹√3 + cot⁻¹(1/√3) = π/3 + π/3 = 2π/3. III = sec⁻¹2 + cosec⁻¹2 = π/3 + π/6 = π/2. Descending order: π > 2π/3 > π/2. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Places π/2 above 2π/3 incorrectly.
Option C → Swaps largest two values.
Option D → Completely reverses order.

🎯 Strategy

Used: Option Grouping

Application:

Compute each expression before comparing.

Final Logic:

Descending order follows π > 2π/3 > π/2.

🧠 Memory Trick

"π biggest, then 120°, then 90°."

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

8 During analytical AC circuit calculations, the structural phase ratio equates perfectly to y = cos⁻¹(cos(7π/6)). Retrieve the exact numerical principal equivalent of y.

A 7π/6

B 5π/6

C π/6

D -π/6

💡 Short Explanation

cos inverse principal range is [0,π]. cos(7π/6)=−√3/2. Principal angle is 5π/6.

✅ Correct Answer Explanation

Since: cos(7π/6)=−√3/2, we evaluate: cos⁻¹(−√3/2). Within principal range [0,π], the corresponding angle is: 5π/6. Thus option B is correct. The original angle 7π/6 lies outside principal range.

❓ Why Other Options Are Incorrect

Option A → Outside principal range.
Option C → Gives positive cosine.
Option D → Negative angle not allowed in cos inverse range.

🎯 Strategy

Used: Elimination

Application:

Check principal range restrictions first.

Final Logic:

cos inverse always outputs angle in [0,π].

🧠 Memory Trick

"cos inverse never negative."

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

9 Match the inverse sine algebraic identity mappings strictly with their exact root equivalents:

List 1	List 2
1. sin⁻¹x = π/2	a. 0
2. sin⁻¹x = −π/2	b. 1/2
3. sin⁻¹x = 0	c. 1
4. sin⁻¹x = π/6	d. −1

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

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

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

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

💡 Short Explanation

sin(π/2)=1. sin(−π/2)=−1. sin(0)=0 and sin(π/6)=1/2.

✅ Correct Answer Explanation

Using standard sine values: sin(π/2)=1, sin(−π/2)=−1, sin0=0, sin(π/6)=1/2. Thus: 1-c, 2-d, 3-a, 4-b. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Swaps π/2 and π/6 mappings.
Option B → Completely mismatched values.
Option D → Incorrect negative-angle association.

🎯 Strategy

Used: Option Grouping

Application:

Convert inverse statements back into direct sine values.

Final Logic:

Use standard unit-circle sine evaluations.

🧠 Memory Trick

"90°→1, 30°→1/2."

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

10 Which formulated analytical statement discussing cos⁻¹x identity domains is strictly incorrect?

A The numerical range covering its principal value branch perfectly rests at [0, π].

B The operation cos⁻¹(cos x) = x asserts validity specifically mapped for x ∈ [0, π].

C The structural function's domain rigorously restricts boundary values strictly to [-1, 1].

D The formal identity states logically that cos⁻¹(-x) = -cos⁻¹x for all valid x values.

💡 Short Explanation

cos inverse range is [0,π]. cos⁻¹(cosx)=x only in [0,π]. cos inverse is not odd.

✅ Correct Answer Explanation

The correct identity is: cos⁻¹(−x)=π−cos⁻¹x, not: −cos⁻¹x. Therefore option D is incorrect. Options A, B, and C are standard NCERT properties of the inverse cosine function and its principal branch.

❓ Why Other Options Are Incorrect

Option A → Correct principal range statement.
Option B → Valid only in [0,π], correctly stated.
Option C → Correct domain restriction for cos inverse.

🎯 Strategy

Used: Odd One Out

Application:

Identify the only identity violating standard inverse cosine properties.

Final Logic:

cos inverse is not an odd function.

🧠 Memory Trick

"cos inverse gives π minus."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Properties of cos⁻¹x

11 A statistician blindly selects input x uniformly mapped from the interval [-1, 1]. What is the explicit probability that substituting x = sin θ forces the principal angle θ to sit completely outside the strict region [-π/4, π/4]?

A 1 - 1/√2

B 1/√2

C 1/2

D 1/4

💡 Short Explanation

θ = sin⁻¹x. Outside [-π/4, π/4] means |x| > 1/√2. Probability = remaining interval length.

✅ Correct Answer Explanation

Since: x = sin θ, and θ belongs to principal branch: [-π/2, π/2]. Condition: θ outside [-π/4, π/4] means: |x| > sin(π/4)=1/√2. Uniform probability on [-1,1]: Required length: 2−2/√2. Probability: (2−2/√2)/2 =1−1/√2. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Represents inside probability, not outside.
Option C → Assumes symmetric half interval incorrectly.
Option D → Ignores sine mapping relation.

🎯 Strategy

Used: Substitution

Application:

Convert angular condition into algebraic sine inequality.

Final Logic:

Outside angular interval corresponds to |x|>1/√2.

🧠 Memory Trick

"π/4 corresponds to 1/√2."

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

12 In the geometric process of simplifying y = tan⁻¹(x) mapped purely on its principal branch, which physical constraint definitively borders its graphical region?

A Fixed vertical asymptotes structured purely at x = ±π/2.

B Rigid horizontal asymptotes mathematically stationed at y = ±π/2.

C Explicit graphical coordinate intercepts at exactly y = ±1.

D A physical discontinuity logically trapped directly at x = 0.

💡 Short Explanation

tan⁻¹x range is (-π/2, π/2). Curve approaches but never touches boundaries. Asymptotes are horizontal.

✅ Correct Answer Explanation

For: y = tan⁻¹x, as: x→∞, y→π/2, and: x→−∞, y→−π/2. Hence the graph has horizontal asymptotes: y=±π/2. Option B correctly describes the graphical boundary behavior of the principal branch.

❓ Why Other Options Are Incorrect

Option A → x=±π/2 are asymptotes of tanx, not tan⁻¹x.
Option C → y=±1 are not asymptotic bounds.
Option D → tan⁻¹x is continuous at x=0.

🎯 Strategy

Used: Elimination

Application:

Differentiate graph properties of tanx and tan⁻¹x.

Final Logic:

Inverse tangent has horizontal—not vertical—asymptotes.

🧠 Memory Trick

"tan inverse flattens near ±90°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Graph of tan⁻¹x –

📄 Passage

We know that the domain of the tan function (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) and so on. The branch with range (-π/2, π/2) is called the principal value branch.

Drawing structurally from the passage, the complete unrestricted domain mapping the function y = tan⁻¹x is established as:

A (-π/2, π/2)

B [-1, 1]

C R

D R - (-1, 1)

💡 Short Explanation

tanx range is R. Inverse domain equals original range. tan⁻¹x accepts all real numbers.

✅ Correct Answer Explanation

The tangent function restricted to: (-π/2, π/2) has range: R. Therefore: tan⁻¹x has: Domain = R, Range = (-π/2, π/2). Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Represents range, not domain.
Option B → Domain belongs to sin inverse/cos inverse.
Option D → Domain of sec inverse/cosec inverse.

🎯 Strategy

Used: Option Grouping

Application:

Match inverse function domain with original function range.

Final Logic:

Inverse tangent accepts every real input.

🧠 Memory Trick

"tan inverse eats all real numbers."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Domain and Range of tan⁻¹x

📄 Passage

Analytically assessing the provided text, why is artificially restricting the tangent function's domain critical prior to establishing tan⁻¹x?

A Because it effectively limits infinite outputs safely to only positive mapping values.

B It physically guarantees calculated discontinuities during algebraic simplification.

C It successfully restricts the continuous curve from crossing the neutral x-axis boundary.

D It mathematically ensures the original function becomes perfectly bijective (one-one and onto), authorizing the inverse mapping.

💡 Short Explanation

Inverse exists only for bijective functions. Restricted tangent becomes one-one and onto. Hence tan⁻¹x becomes well-defined.

✅ Correct Answer Explanation

The unrestricted tangent function is periodic and not one-one. Restricting its domain to: (-π/2, π/2) makes it bijective: both one-one and onto. Only then can the inverse function tan⁻¹x be defined properly. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Positivity is unrelated to invertibility.
Option B → Restriction does not create discontinuity.
Option C → Crossing x-axis is irrelevant.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Identify the NCERT definition of invertibility.

Final Logic:

Inverse exists only when function becomes bijective.

🧠 Memory Trick

"Restrict → biject → inverse."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Need for Restricted Domains –

15 Algebraically reduce and formally simplify the structural expression tan⁻¹((cos x - sin x) / (cos x + sin x)), assuming strictly x < π.

A π/4 - x

B π/4 + x

C x - π/4

D π/2 - x

💡 Short Explanation

Divide numerator and denominator by cosx. Use tangent subtraction identity. tan(π/4−x) obtained.

✅ Correct Answer Explanation

Expression: (cosx−sinx)/(cosx+sinx) Divide by cosx: (1−tanx)/(1+tanx). Using identity: tan(A−B)=(tanA−tanB)/(1+tanAtanB), with: A=π/4, we get: tan(π/4−x). Therefore: tan⁻¹(...) = π/4−x.

❓ Why Other Options Are Incorrect

Option B → Wrong sign in tangent identity.
Option C → Reversed expression.
Option D → Not generated by subtraction formula.

🎯 Strategy

Used: Substitution

Application:

Rewrite expression into tangent identity form.

Final Logic:

Expression matches tan(π/4−x).

🧠 Memory Trick

"1−tan over 1+tan means π/4 minus."

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

16 Map the rational format expression cot⁻¹(1 / √(x² - 1)) carefully into its most basic identical simplified state exactly where x > 1.

A cosec⁻¹x

B sec⁻¹x

C sin⁻¹x

D cos⁻¹x

💡 Short Explanation

Let x=secθ. Then √(x²−1)=tanθ. cot⁻¹(cotθ)=θ=sec⁻¹x.

✅ Correct Answer Explanation

For: x>1, let: x=secθ. Then: √(x²−1)=tanθ. So: 1/√(x²−1)=cotθ. Hence: cot⁻¹(cotθ)=θ=sec⁻¹x. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Corresponds to cosecant substitution.
Option C → Uses sine relation incorrectly.
Option D → No cosine inverse form appears.

🎯 Strategy

Used: Substitution

Application:

Introduce standard secant substitution.

Final Logic:

Expression naturally converts into sec inverse.

🧠 Memory Trick

"√(x²−1) signals secθ substitution."

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

17 Analytically break down and evaluate the theoretical trigonometric combination mapping sin(tan⁻¹ x), assuming boundaries bounded securely by |x| < 1.

A x / √(1-x²)

B 1 / √(1+x²)

C x / √(1+x²)

D √(1+x²) / x

💡 Short Explanation

Let θ=tan⁻¹x. tanθ=x/1. Use triangle relation for sine.

✅ Correct Answer Explanation

Let: θ=tan⁻¹x. Then: tanθ=x. Take right triangle: Opposite=x, Adjacent=1. Hypotenuse: √(1+x²). Therefore: sinθ = x/√(1+x²). Hence: sin(tan⁻¹x)=x/√(1+x²).

❓ Why Other Options Are Incorrect

Option A → Corresponds to sine substitution form.
Option B → Gives cosine value instead.
Option D → Reciprocal expression incorrect.

🎯 Strategy

Used: Substitution

Application:

Convert inverse tangent into triangle ratio.

Final Logic:

Sine = opposite/hypotenuse.

🧠 Memory Trick

"tan inverse → x,1 triangle."

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

18 The theoretical trigonometric identity equation sin⁻¹(1-x) - 2sin⁻¹x = π/2 holds structurally true strictly for which explicit subset boundary value(s) of x?

A 1, 1/2

B 1/2

C 0, 1/2

D 0

💡 Short Explanation

Test boundary values directly. x=0 satisfies identity. x=1/2 does not satisfy equality.

✅ Correct Answer Explanation

Check: x=0. LHS: sin⁻¹(1)−2sin⁻¹0 =π/2−0 =π/2. Hence true. For: x=1/2, LHS: sin⁻¹(1/2)−2sin⁻¹(1/2) =π/6−π/3 =−π/6. Not equal to π/2. Thus only x=0 works.

❓ Why Other Options Are Incorrect

Option A → x=1/2 invalid.
Option B → Gives incorrect equality.
Option C → Contains extra incorrect value.

🎯 Strategy

Used: Substitution

Application:

Directly test listed values.

Final Logic:

Only x=0 satisfies equation.

🧠 Memory Trick

"Always test boundary values first."

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

19 Locate exactly the correct principal value derived systematically from evaluating cos⁻¹(-1/2) in the practice module.

A 2π/3

B π/3

C 5π/6

D -π/3

💡 Short Explanation

cos inverse range is [0,π]. cos(2π/3)=−1/2. Principal angle chosen in valid range.

✅ Correct Answer Explanation

We need angle θ in: [0,π] such that: cosθ=−1/2. This occurs at: θ=2π/3. Hence: cos⁻¹(−1/2)=2π/3. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Gives positive cosine.
Option C → cosine equals −√3/2 there.
Option D → Negative angle outside principal range.

🎯 Strategy

Used: Elimination

Application:

Apply principal range restriction for cosine inverse.

Final Logic:

Only 2π/3 gives cosine −1/2 in [0,π].

🧠 Memory Trick

"Negative cosine lives in second quadrant."

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

20 Verify precisely the final principal numerical boundary output mapped structurally for tan⁻¹(-1).

A π/4

B -π/4

C 3π/4

D 5π/4

💡 Short Explanation

tan inverse range is (-π/2, π/2). tan(−π/4)=−1. Principal value must stay in range.

✅ Correct Answer Explanation

The principal range of tan⁻¹x is: (−π/2, π/2). We require angle θ such that: tanθ=−1. Inside the principal range: θ=−π/4. Thus: tan⁻¹(−1)=−π/4. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Gives positive tangent.
Option C → Outside principal range.
Option D → Also outside principal range.

🎯 Strategy

Used: Elimination

Application:

Check tangent sign and principal branch simultaneously.

Final Logic:

Only −π/4 satisfies both conditions.

🧠 Memory Trick

"Negative tangent in principal branch → −45°."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Principal Values of tan⁻¹x

CUET UG Mathematics Booster Test 2 - Applications and Identities of Inverse Trigonometric Functions

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