CUET UG Mathematics Booster Test 1 - Advanced Applications of Inverse Trigonometric Functions -

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CUET UG Mathematics Booster Test 1 - Advanced Applications 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 1 - Advanced Applications of Inverse Trigonometric Functions

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QUESTION 1 OF 20

Assertion (A): sec⁻¹(2) exists and its principal value is π/3.
Reason (R): The domain of sec⁻¹x is (-∞,-1] ∪ [1,∞), and its principal value lies in [0,π], excluding π/2.

Answer Review

1 Assertion (A): sec⁻¹(2) exists and its principal value is π/3.
Reason (R): The domain of sec⁻¹x is (-∞,-1] ∪ [1,∞), and its principal value lies in [0,π], excluding π/2.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

sec⁻¹x exists for |x| ≥ 1. sec(π/3)=2. Principal range excludes π/2.

✅ Correct Answer Explanation

The function sec⁻¹x is defined for x ≤ -1 or x ≥ 1. Since 2 belongs to this domain, sec⁻¹(2) exists. Also, sec(π/3)=2 and π/3 lies in the principal value branch [0,π] excluding π/2. Therefore both Assertion and Reason are true, and the Reason correctly explains the Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct.
Option B → The reason is true because it correctly states domain and principal range.
Option D → Assertion is also true since sec⁻¹(2)=π/3.

🎯 Strategy

Used: Substitution

Application:

Substitute θ=π/3 into secθ and verify domain conditions.

Final Logic:

sec(π/3)=2 and π/3 belongs to the principal branch.

🧠 Memory Trick

"sec⁻¹ accepts numbers outside ±1."

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

2 A force vector makes an angle θ such that
cot⁻¹(1)=θ.
Find the principal value of θ.

A π/4

B π/2

C 3π/4

D 0

💡 Short Explanation

cot(π/4)=1. Principal range is (0,π). Hence θ=π/4.

✅ Correct Answer Explanation

The principal value branch of cot⁻¹x lies in (0,π). Since cot(π/4)=1, the principal value satisfying cot⁻¹(1)=θ is π/4. The angle π/4 lies within the valid principal range, making option A correct.

❓ Why Other Options Are Incorrect

Option B → cot(π/2)=0, not 1.
Option C → cot(3π/4)=-1, not 1.
Option D → cot(0) is undefined.

🎯 Strategy

Used: Substitution

Application:

Use standard trigonometric values inside principal range.

Final Logic:

cot(π/4)=1 uniquely determines θ.

🧠 Memory Trick

"cot 45° equals 1."

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

3 The domain of sin⁻¹x is [-1,1]. A value is chosen uniformly at random from this domain.
What is the probability that the principal value lies in (0,π/2)?

A 1/4

B 1/2

C 3/4

D 1/2

💡 Short Explanation

Positive x gives positive sin⁻¹x. Half interval is positive. Probability equals 1/2.

✅ Correct Answer Explanation

The principal range of sin⁻¹x is [-π/2,π/2]. The output lies in (0,π/2) whenever x∈(0,1]. From the domain [-1,1], exactly half the interval corresponds to positive x-values. Hence the probability is 1/2.

❓ Why Other Options Are Incorrect

Option A → Underestimates favorable region.
Option C → Overestimates positive interval size.
Option D → Duplicate numerical value appears, but B is intended correct choice.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate positivity of sin⁻¹x directly to positivity of x.

Final Logic:

Positive half of the domain gives positive principal values.

🧠 Memory Trick

"sin⁻¹ keeps the sign of x."

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

4 Evaluate:
tan(sin⁻¹(1/2))

A 1/√3

B 1/√3

C √3

D 1

💡 Short Explanation

sin⁻¹(1/2)=π/6. tan(π/6)=1/√3. Standard angle evaluation.

✅ Correct Answer Explanation

First evaluate: sin⁻¹(1/2)=π/6 because π/6 lies in the principal range of sin⁻¹x. Now: tan(π/6)=1/√3. Hence the expression equals 1/√3, making option A correct.

❓ Why Other Options Are Incorrect

Option B → Duplicate of option A.
Option C → tan(π/3)=√3, unrelated here.
Option D → tan(π/4)=1, not π/6.

🎯 Strategy

Used: Substitution

Application:

Convert inverse trig value into standard angle.

Final Logic:

sin⁻¹(1/2)=π/6 leads directly to tan(π/6).

🧠 Memory Trick

"Half gives 30°, tan30°=1/√3."

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

5 Arrange in ascending order:
1. sin(sin⁻¹0.2)
2. sin(sin⁻¹0.5)
3. sin(sin⁻¹0.8)
4. sin(sin⁻¹0.3)

A 1, 2, 3, 4

B 1, 4, 2, 3

C 3, 1, 4, 2

D 2, 4, 3, 1

💡 Short Explanation

sin(sin⁻¹x)=x. Values become 0.2, 0.5, 0.8, 0.3. Arrange numerically.

✅ Correct Answer Explanation

Using the identity: sin(sin⁻¹x)=x. The values become: 0.2, 0.5, 0.8, 0.3. Ascending order: 0.2 < 0.3 < 0.5 < 0.8. Therefore: 1,4,2,3. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Places 0.8 before 0.3 incorrectly.
Option C → Starts with the largest value first.
Option D → Entire sequence is not ascending.

🎯 Strategy

Used: Substitution

Application:

Reduce composite functions to direct numerical values.

Final Logic:

Arrange simplified values in increasing order.

🧠 Memory Trick

"sin and sin⁻¹ cancel."

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

6 The integral calculation of a planar region involving inverse trigonometric functions requires using their principal value ranges. What is the principal value range of the function sin⁻¹x?

A (-π,π)

B [0,π]

C (-π/2,π/2)

D [-π/2,π/2]

💡 Short Explanation

sin⁻¹x requires bijective sine branch. Standard branch is symmetric. Endpoints included.

✅ Correct Answer Explanation

To define inverse sine uniquely, the sine function is restricted to: [-π/2,π/2]. Within this interval sine is one-one and onto [-1,1]. Hence the principal value range of sin⁻¹x is exactly [-π/2,π/2], making option D correct.

❓ Why Other Options Are Incorrect

Option A → Too wide; sine repeats values there.
Option B → Corresponds to cos⁻¹x range.
Option C → Incorrect because endpoints ±π/2 are included.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall the standard principal branch used for inverse sine.

Final Logic:

Inverse sine uses the restricted interval [-π/2,π/2].

🧠 Memory Trick

"sin⁻¹ lives between ±90°."

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

7 Which of the following intervals are NOT standard principal value branches for inverse trigonometric functions?
1. [-π/2,π/2]
2. [0,π]
3. (-π,π)
4. [0,2π]

A 2, 3, and 4 only

B 1 and 2 only

C 1, 3, and 4 only

D 3 and 4 only

💡 Short Explanation

Standard branches include intervals 1 and 2. (-π,π) and [0,2π] are not standard. Hence 3 and 4 only.

✅ Correct Answer Explanation

Standard principal value branches are: [-π/2,π/2] for sin⁻¹x and tan⁻¹x, and [0,π] for cos⁻¹x. Intervals (-π,π) and [0,2π] are not standard principal branches because trigonometric functions are not one-one there. Therefore option D is correct. The provided answer A is incorrect.

❓ Why Other Options Are Incorrect

Option A → Incorrectly includes [0,π], which is a valid standard branch.
Option B → Both intervals listed are standard principal branches.
Option C → Incorrectly marks [-π/2,π/2] as non-standard.

🎯 Strategy

Used: Elimination

Application:

Identify intervals officially used in NCERT principal branches.

Final Logic:

Only intervals 3 and 4 fail as standard branches.

🧠 Memory Trick

"Principal branches avoid repeated values."

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

8 Match each trigonometric function with the domain restriction required to make it bijective (for defining its inverse):

List 1	List 2
1. Sine	a. (0, π)
2. Cosine	b. [−π/2, π/2]
3. Tangent	c. (−π/2, π/2)
4. Cotangent	d. (0, π)

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

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

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

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

💡 Short Explanation

Sine uses [-π/2,π/2]. Cosine and cotangent use (0,π). Tangent uses (-π/2,π/2).

✅ Correct Answer Explanation

Standard bijective restrictions are: Sine → [-π/2,π/2] Cosine → [0,π] or equivalent convention near (0,π) Tangent → (-π/2,π/2) Cotangent → (0,π). Thus matching corresponds best with option B. The provided answer C is incorrect because cosine does not use (-π/2,π/2).

❓ Why Other Options Are Incorrect

Option A → Incorrectly assigns sine to (0,π).
Option C → Gives cosine wrong interval.
Option D → Assigns sine incorrect tangent-style interval.

🎯 Strategy

Used: Option Grouping

Application:

Recall standard restricted intervals used for each inverse function.

Final Logic:

Only option B correctly matches all principal restrictions.

🧠 Memory Trick

"sin symmetric, tan open, cos and cot positive π interval."

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

9 Consider the equation:
sinx/(1-cosx)=tan(x/2)
Which of the following statements is INCORRECT?

A The equation has a valid solution at x=π/2

B The expression can be derived using standard identities

C cosx ≠ 1 to avoid division by zero

D The equation is valid for all real x

💡 Short Explanation

Denominator cannot be zero. cosx=1 causes undefined form. Hence not valid for all real x.

✅ Correct Answer Explanation

Using half-angle identities: sinx/(1-cosx)=cot(x/2), while related identities involve tan(x/2). The expression becomes undefined whenever: 1-cosx=0 ⇒ cosx=1. Thus it cannot hold for all real x. Hence option D is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → x=π/2 gives valid finite values.
Option B → Such relations arise from standard trigonometric identities.
Option C → Correct restriction avoids zero denominator.

🎯 Strategy

Used: Extreme Word Filter

Application:

The phrase "all real x" often signals invalid overgeneralization.

Final Logic:

Expressions with denominators always need restrictions.

🧠 Memory Trick

"Denominator zero destroys identity."

📚 Class XI – NCERT Mathematics – Chapter: Trigonometric Functions – Topic: Half-Angle Identities

10 Consider the relation:
y = tan⁻¹x
What are the horizontal asymptotes of this graph?

A y=0 and y=π

B y=-π/2 and y=π/2

C y=-π and y=π

D No horizontal asymptotes

💡 Short Explanation

tan⁻¹x principal range is bounded. As x→±∞, outputs approach ±π/2. Hence horizontal asymptotes exist.

✅ Correct Answer Explanation

For the inverse tangent function: y=tan⁻¹x, as: x→∞, y→π/2 and x→−∞, y→−π/2. Therefore the graph approaches the horizontal lines: y=±π/2. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Corresponds to cot⁻¹x-type limits.
Option C → tan⁻¹x never approaches ±π.
Option D → Horizontal asymptotes clearly exist.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall standard graph behavior of inverse tangent.

Final Logic:

tan⁻¹x is bounded between ±π/2.

🧠 Memory Trick

"tan⁻¹ hugs ±90°."

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

11 Given the angles sin⁻¹(1/2), cos⁻¹(1/2), and tan⁻¹(1), find their mean value.

A π/4

B π/3

C π/3

D π/2

💡 Short Explanation

sin⁻¹(1/2)=π/6. cos⁻¹(1/2)=π/3 and tan⁻¹(1)=π/4. Average equals π/4.

✅ Correct Answer Explanation

The principal values are: sin⁻¹(1/2)=π/6, cos⁻¹(1/2)=π/3, tan⁻¹(1)=π/4. Their mean is: (π/6 + π/3 + π/4)/3 = (2π+4π+3π)/12 ÷3 = 9π/12 ÷3 = 3π/4 ÷3 = π/4. Hence option A is correct. The provided explanation incorrectly simplified to π/3.

❓ Why Other Options Are Incorrect

Option B → Arithmetic simplification error.
Option C → Duplicate incorrect option.
Option D → Too large compared to actual average.

🎯 Strategy

Used: Substitution

Application:

Replace each inverse trigonometric value with its standard angle.

Final Logic:

Correct averaging gives π/4.

🧠 Memory Trick

"30°, 60°, 45° average to 45°."

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

12 Evaluate the area represented by:
∫₀¹ 1/√(1-x²) dx

A π/2

B π

C 1

D 0

💡 Short Explanation

Integral equals sin⁻¹x. Evaluate from 0 to 1. Result is π/2.

✅ Correct Answer Explanation

Using the standard integration formula: ∫ dx/√(1-x²)=sin⁻¹x + C. Evaluating between 0 and 1: sin⁻¹(1)-sin⁻¹(0) = π/2 - 0 = π/2. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Double the correct value.
Option C → Ignores inverse trigonometric evaluation.
Option D → Integral over positive interval cannot vanish.

🎯 Strategy

Used: Substitution

Application:

Recognize standard inverse trigonometric integral form.

Final Logic:

Boundary substitution directly yields π/2.

🧠 Memory Trick

"1/√(1-x²) → sin⁻¹x."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Standard Integrals

13 The domain of inverse trigonometric functions collectively spans:

A All real numbers

B [-1,1]

C (0,π)

D (-π/2,π/2)

💡 Short Explanation

Different inverse functions have different domains. tan⁻¹x and cot⁻¹x accept all reals. Collectively domains cover R.

✅ Correct Answer Explanation

Inverse trigonometric functions do not share one common domain. Functions like tan⁻¹x and cot⁻¹x are defined for all real numbers, while sin⁻¹x and cos⁻¹x are restricted to [-1,1]. Hence, taken collectively, inverse trigonometric domains span all real numbers. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Only for sin⁻¹x and cos⁻¹x.
Option C → Represents principal range of cot⁻¹x or cos⁻¹x.
Option D → Principal range, not collective domain.

🎯 Strategy

Used: Option Grouping

Application:

Compare domains of all inverse trigonometric functions together.

Final Logic:

tan⁻¹x extends the collective domain to all reals.

🧠 Memory Trick

"tan⁻¹ unlocks all real numbers."

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

14 What is the principal value branch of cos⁻¹x?

A [-π/2,π/2]

B (-π,π)

C (0,2π)

D [0,π]

💡 Short Explanation

cos⁻¹x requires one-one cosine branch. Standard restriction is [0,π]. Endpoints included.

✅ Correct Answer Explanation

To define cos⁻¹x uniquely, cosine is restricted to [0,π], where it becomes bijective. Therefore the principal value branch of cos⁻¹x is [0,π]. This interval includes both endpoints and covers the entire range of cosine from 1 to -1.

❓ Why Other Options Are Incorrect

Option A → Principal branch of sin⁻¹x.
Option B → Cosine repeats values in this interval.
Option C → Too large; cosine not one-one there.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall standard NCERT principal branch definitions.

Final Logic:

cos⁻¹x always uses [0,π].

🧠 Memory Trick

"Cos travels from 0° to 180°."

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

15 Why must we define principal value branches for inverse trigonometric functions?

A To make the curves continuously differentiable everywhere

B Because trigonometric functions are not one–one over their full domains

C To simplify probability calculations

D To follow classical geometry conventions

💡 Short Explanation

Inverses require one-one mappings. Trigonometric functions are periodic. Restrictions create bijective branches.

✅ Correct Answer Explanation

Inverse functions exist only when the original function is one-one and onto. Trigonometric functions repeat values periodically over their natural domains, so they are not one-one. Hence principal value branches are defined by restricting domains appropriately. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Differentiability is not the main reason.
Option C → Probability has no role in branch definition.
Option D → Mathematical necessity, not convention alone.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the basic inverse-function condition of injectivity.

Final Logic:

Periodic repetition prevents unrestricted inverses.

🧠 Memory Trick

"No one-one, no inverse."

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

16 Evaluate using principal value branches:
sin⁻¹(1/2)+cos⁻¹(1/2)

A π/6

B π/3

C π/2

D π

💡 Short Explanation

sin⁻¹(1/2)=π/6. cos⁻¹(1/2)=π/3. Sum equals π/2.

✅ Correct Answer Explanation

Using principal values: sin⁻¹(1/2)=π/6 and cos⁻¹(1/2)=π/3. Adding: π/6 + π/3 = π/6 + 2π/6 = 3π/6 = π/2. Hence option C is correct. This also follows the identity: sin⁻¹x + cos⁻¹x = π/2.

❓ Why Other Options Are Incorrect

Option A → Only equals sin⁻¹(1/2).
Option B → Only equals cos⁻¹(1/2).
Option D → Sum is much smaller than π.

🎯 Strategy

Used: Substitution

Application:

Replace inverse values with standard angles.

Final Logic:

Known identity gives π/2 immediately.

🧠 Memory Trick

"sin⁻¹x + cos⁻¹x = 90°."

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

📄 Passage

"The study of trigonometry was first started in India. The ancient Indian Mathematicians, Aryabhata (476 A.D.), Brahmagupta (598 A.D.), Bhaskara I (600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. All this knowledge went from India to Arabia and then from there to Europe. The Greeks had also started the study of trigonometry but their approach was so clumsy that when the Indian approach became known, it was immediately adopted throughout the world."

According to the passage, the Indian approach to trigonometry was adopted worldwide because:

A It was exclusively translated into Arabic texts

B Europe mandated a new mathematical system

C Aryabhata enforced its global use

D The Greek approach to trigonometry was considered clumsy

💡 Short Explanation

Passage compares Greek and Indian approaches. Greek method called clumsy. Indian system became preferred worldwide.

✅ Correct Answer Explanation

The passage explicitly states that the Greek approach to trigonometry was "clumsy." When the Indian approach became known, it was adopted worldwide because it was more effective and systematic. Hence option D directly matches the passage statement and is correct.

❓ Why Other Options Are Incorrect

Option A → Translation alone was not the reason stated.
Option B → No European mandate is mentioned.
Option C → Aryabhata did not enforce global adoption.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Identify the exact textual reason given in the passage.

Final Logic:

The passage directly attributes adoption to Greek inefficiency.

🧠 Memory Trick

"Clumsy Greeks, smooth Indian method."

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Historical Development of Trigonometry

📄 Passage

Arrange the following mathematicians chronologically based on their contributions:
1. Bhaskara II
2. Aryabhata
3. Brahmagupta
4. Bhaskara I

A 2, 3, 4, 1

B 1, 2, 3, 4

C 3, 2, 1, 4

D 2, 4, 3, 1

💡 Short Explanation

Aryabhata: 476 A.D. Brahmagupta: 598 A.D. Bhaskara I: 600 A.D.; Bhaskara II: 1114 A.D.

✅ Correct Answer Explanation

Chronological order by year: Aryabhata (476 A.D.) → Brahmagupta (598 A.D.) → Bhaskara I (600 A.D.) → Bhaskara II (1114 A.D.). Thus the correct sequence is: 2,3,4,1. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Begins with latest mathematician first.
Option C → Places Brahmagupta before Aryabhata incorrectly.
Option D → Swaps Brahmagupta and Bhaskara I incorrectly.

🎯 Strategy

Used: Arrange Chronologically

Application:

Use historical years given in the passage.

Final Logic:

Ascending year order determines sequence.

🧠 Memory Trick

"Arya → Brahma → Bhaskara I → Bhaskara II."

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

19 Aryabhata is historically recorded to have made his major mathematical contributions around which year?

A 598 A.D.

B 600 A.D.

C 476 A.D.

D 1114 A.D.

💡 Short Explanation

Aryabhata lived earliest among listed scholars. Passage explicitly gives 476 A.D. Historical fact-based question.

✅ Correct Answer Explanation

The passage directly states: "Aryabhata (476 A.D.)." Therefore his major mathematical contributions are associated with 476 A.D. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Corresponds to Brahmagupta.
Option B → Corresponds to Bhaskara I.
Option D → Corresponds to Bhaskara II.

🎯 Strategy

Used: Elimination

Application:

Match mathematicians with their historically associated years.

Final Logic:

Only 476 A.D. belongs to Aryabhata.

🧠 Memory Trick

"Aryabhata → 476."

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

20 The 16th-century Malayalam mathematical work Yuktibhasa contains a detailed proof for the infinite series expansion of:

A sinx = x - x³/3! + x⁵/5! - ⋯

B tan⁻¹x = x - x³/3 + x⁵/5 - ⋯

C eˣ = 1 + x + x²/2! + ⋯

D cosx = 1 - x²/2! + x⁴/4! - ⋯

💡 Short Explanation

Yuktibhasa discussed infinite series. Famous result includes arctangent expansion. Linked to Kerala School mathematics.

✅ Correct Answer Explanation

The Kerala School text Yuktibhasa is historically known for rigorous derivations of infinite series expansions, including: tan⁻¹x = x − x³/3 + x⁵/5 − ⋯ This series later became famous in calculus developments. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Sine series is famous but not the highlighted result here.
Option C → Exponential series belongs to later analysis developments.
Option D → Cosine series is not the intended historical reference.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall Kerala School contribution to inverse tangent series.

Final Logic:

Yuktibhasa is strongly associated with arctangent expansion.

🧠 Memory Trick

"Yuktibhasa → tan⁻¹ series."

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

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