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

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

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QUESTION 1 OF 20

Evaluate the differences carefully. Which option correctly identifies multiple valid distinctions between sin⁻¹x and (sin x)⁻¹?

Answer Review

1 Evaluate the differences carefully. Which option correctly identifies multiple valid distinctions between sin⁻¹x and (sin x)⁻¹?

A Both mathematical notations represent the arcsine function perfectly.

B sin⁻¹x is strictly the inverse angle function, whereas (sin x)⁻¹ represents the cosecant function.

C They explicitly share the exact same mathematical domain [-1, 1].

D Both functions are properly defined at the point x=0.

💡 Short Explanation

sin⁻¹x means inverse sine function. (sinx)⁻¹ means reciprocal of sinx. Reciprocal equals cosec x.

✅ Correct Answer Explanation

The notation sin⁻¹x denotes the inverse sine function (arcsine), whereas: (sinx)⁻¹ = 1/sinx = cosec x. Thus option B correctly distinguishes inverse and reciprocal meanings. Option A incorrectly equates them. Option C is false because their domains differ. Option D is false since cosec x is undefined at x=0.

❓ Why Other Options Are Incorrect

Option A → Inverse function and reciprocal function are conceptually different.
Option C → sin⁻¹x has domain [-1,1], while cosec x excludes multiples of π.
Option D → cosec0 is undefined because sin0=0.

🎯 Strategy

Used: Odd One Out

Application:

Identify the only option correctly distinguishing inverse and reciprocal notation.

Final Logic:

Inverse sine and reciprocal sine are different mathematical functions.

🧠 Memory Trick

"−1 outside bracket means inverse; inside means reciprocal."

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

2 An electrical engineer calculates a current's phase angle y = sin⁻¹(-1/√2) as a principal value. Find this exact numerical angle in radians.

A π/4

B 3π/4

C -π/4

D -3π/4

💡 Short Explanation

sin(−π/4)=−1/√2. Principal branch is [−π/2, π/2]. Hence principal value is −π/4.

✅ Correct Answer Explanation

The principal range of sin⁻¹x is: [−π/2, π/2]. Since: sin(−π/4)=−1/√2, the principal value becomes: sin⁻¹(−1/√2)=−π/4. Thus option C is correct. Options B and D lie outside the principal branch or produce incorrect sine values.

❓ Why Other Options Are Incorrect

Option A → Gives positive sine value.
Option B → Lies outside principal branch.
Option D → sine equals +1/√2 there.

🎯 Strategy

Used: Substitution

Application:

Recall standard unit-circle inverse sine values.

Final Logic:

Inverse sine returns the principal angle inside [−π/2, π/2].

🧠 Memory Trick

"Negative input → negative principal sine angle."

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

3 Assertion (A): sin⁻¹(sin(3π/5)) strictly equals 3π/5.
Reason (R): The formula sin⁻¹(sin x) = x is valid exclusively for x ∈ [-π/2, π/2].

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

3π/5 lies outside principal branch. sin⁻¹(sinx)=x only inside branch. Correct value becomes 2π/5.

✅ Correct Answer Explanation

The identity: sin⁻¹(sinx)=x holds only for: x ∈ [−π/2, π/2]. Since: 3π/5 > π/2, principal adjustment is required: sin⁻¹(sin3π/5)=2π/5. Thus assertion is false, but the reason is true. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason statement is correct.
Option B → Assertion is false.
Option C → Assertion itself is incorrect.

🎯 Strategy

Used: Elimination

Application:

Check whether the angle lies inside principal branch.

Final Logic:

Outside principal branch, inverse sine requires angle correction.

🧠 Memory Trick

"Outside branch ⇒ adjust angle."

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

4 A phase data set contains values D = {cot⁻¹(-1), sec⁻¹(-2)}. Calculate the moving average (mean) of these two principal values.

A 17π/24

B 17π/12

C 5π/12

D 7π/24

💡 Short Explanation

cot⁻¹(−1)=3π/4. sec⁻¹(−2)=2π/3. Mean equals 17π/24.

✅ Correct Answer Explanation

Using principal values: cot⁻¹(−1)=3π/4 and: sec⁻¹(−2)=2π/3. Mean: [(3π/4)+(2π/3)]/2 = [9π/12+8π/12]/2 =17π/24. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Sum taken without averaging.
Option C → Arithmetic mistake in addition.
Option D → Uses incorrect principal values.

🎯 Strategy

Used: Substitution

Application:

Replace inverse functions with standard exact values.

Final Logic:

Average = (sum of principal values)/2.

🧠 Memory Trick

"−1 cot → 135°, −2 sec → 120°."

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

5 Evaluate the definite mathematical integral of f(x) = sec⁻¹(2) - cosec⁻¹(√2) mapped strictly from x=0 to x=2.

A π/12

B π/6

C π/3

D π/4

💡 Short Explanation

sec⁻¹2=π/3. cosec⁻¹√2=π/4. Constant integral gives 2×π/12=π/6.

✅ Correct Answer Explanation

Evaluate the constant: sec⁻¹2−cosec⁻¹√2 =π/3−π/4 =π/12. Integral from 0 to 2: ∫₀² (π/12)dx =(π/12)(2) =π/6. Thus option B is correct.

❓ Why Other Options Are Incorrect

Option A → Only evaluates constant difference.
Option C → Incorrect multiplication.
Option D → Wrong principal value substitution.

🎯 Strategy

Used: Substitution

Application:

Evaluate inverse values first, then integrate constant.

Final Logic:

Constant integral = constant × interval length.

🧠 Memory Trick

"Integrate constant? Multiply by interval."

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

6 Calculate the enclosed area of a geometric right triangle featuring a base equal to cos⁻¹(1/2) and a height equal to sin⁻¹(√3/2).

A π²/6

B π²/12

C π²/18

D π²/9

💡 Short Explanation

cos⁻¹(1/2)=π/3. sin⁻¹(√3/2)=π/3. Area=(1/2)(π/3)(π/3).

✅ Correct Answer Explanation

Using principal values: cos⁻¹(1/2)=π/3 and: sin⁻¹(√3/2)=π/3. Triangle area: ½ × (π/3) × (π/3) = π²/18. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Misses triangle half factor.
Option B → Arithmetic simplification error.
Option D → Computes rectangle area instead.

🎯 Strategy

Used: Substitution

Application:

Convert inverse expressions into exact angles.

Final Logic:

Triangle area formula gives π²/18.

🧠 Memory Trick

"π/3 base and height → divide by 2."

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

7 Arrange the following mathematical sums in sequential ascending order:
I) sin⁻¹(0)+cos⁻¹(1),
II) tan⁻¹(1)+cot⁻¹(1),
III) sec⁻¹(√2)+cosec⁻¹(1).

A III, II, I

B II, I, III

C I, III, II

D I, II, III

💡 Short Explanation

I=0+0=0. II=π/4+π/4=π/2. III=π/4+π/2=3π/4.

✅ Correct Answer Explanation

Evaluate each: I = sin⁻¹0 + cos⁻¹1 = 0 + 0 = 0. II = tan⁻¹1 + cot⁻¹1 = π/4 + π/4 = π/2. III = sec⁻¹√2 + cosec⁻¹1 = π/4 + π/2 = 3π/4. Ascending order: 0 < π/2 < 3π/4. Thus option D is correct.

❓ Why Other Options Are Incorrect

Option A → Completely reverses correct order.
Option B → Places π/2 before 0 incorrectly.
Option C → Swaps second and third values.

🎯 Strategy

Used: Option Grouping

Application:

Evaluate each sum numerically, then compare.

Final Logic:

0 < π/2 < 3π/4.

🧠 Memory Trick

"0, 90°, 135°."

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

8 Two resultant vectors point at distinct direction angles α = tan⁻¹(1) and β = cos⁻¹(0). Determine the positive angle difference |α - β| between them.

A π/4

B π/2

C π/3

D π/6

💡 Short Explanation

tan⁻¹1=π/4. cos⁻¹0=π/2. Difference equals π/4.

✅ Correct Answer Explanation

Using standard principal values: tan⁻¹1=π/4 and: cos⁻¹0=π/2. Hence: |α−β| = |π/4−π/2| = π/4. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Gives larger angle directly.
Option C → No correct subtraction.
Option D → Half of actual difference.

🎯 Strategy

Used: Substitution

Application:

Replace inverse expressions with exact standard angles.

Final Logic:

Absolute difference equals π/4.

🧠 Memory Trick

"45° and 90° differ by 45°."

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

9 Match the inverse expressions strictly with their simplest identity forms (assuming |x| < 1):

List 1	List 2
1. sin(sin⁻¹x)	a. π/4
2. sin⁻¹(sin π/4)	b. sin⁻¹(2x√(1 − x²))
3. 2sin⁻¹x	c. x
4. sin⁻¹(−x)	d. −sin⁻¹x

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

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

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

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

💡 Short Explanation

sin(sin⁻¹x)=x. sin⁻¹(sinπ/4)=π/4. Use double-angle identity.

✅ Correct Answer Explanation

Using standard identities: 1. sin(sin⁻¹x)=x. 2. sin⁻¹(sinπ/4)=π/4 because π/4 lies in principal branch. 3. 2sin⁻¹x = sin⁻¹(2x√(1−x²)) under restrictions. 4. sin⁻¹(−x)=−sin⁻¹x. Thus correct matching is: 1-c, 2-a, 3-b, 4-d.

❓ Why Other Options Are Incorrect

Option A → Swaps first two identities incorrectly.
Option C → Incorrectly maps second identity.
Option D → Wrongly assigns odd-function property.

🎯 Strategy

Used: Option Grouping

Application:

Match each identity using standard NCERT formulas.

Final Logic:

All identities align exactly with option B.

🧠 Memory Trick

"sin inverse is odd."

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

10 For the inverse identity curve y = cos⁻¹x, identify the correct graphical region characteristic mapped over its domain [-1, 1].

A It passes straight through the origin (0,0).

B The function behaves as a strictly increasing curve.

C It represents a strictly decreasing curve passing through the intercept (0, π/2).

D The geometric curve is perfectly symmetric about the y-axis.

💡 Short Explanation

cos⁻¹x decreases on [−1,1]. cos⁻¹0=π/2. Curve crosses y-axis at π/2.

✅ Correct Answer Explanation

The function: y=cos⁻¹x is strictly decreasing over: [−1,1]. Also: cos⁻¹0=π/2, so the graph passes through: (0,π/2). Hence option C is correct. The graph neither passes through origin nor remains symmetric about the y-axis.

❓ Why Other Options Are Incorrect

Option A → cos⁻¹0 is π/2, not 0.
Option B → cos inverse decreases, not increases.
Option D → Function lacks y-axis symmetry.

🎯 Strategy

Used: Elimination

Application:

Use monotonicity and intercept properties of cos inverse graph.

Final Logic:

cos⁻¹x decreases from π to 0 across domain.

🧠 Memory Trick

"cos inverse falls from π to 0."

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

11 If an analytical angle θ is uniformly distributed strictly in [0, π], what is the probability that θ effectively lies in the principal value branch of sec⁻¹x?

A 0

B 1/2

C π/2

D 1

💡 Short Explanation

Principal branch of sec⁻¹ is [0, π] − {π/2}. Single excluded point has zero probability. Entire interval contributes probability 1.

✅ Correct Answer Explanation

The principal range of sec⁻¹x is: [0, π] − {π/2}. If θ is uniformly distributed over [0, π], excluding a single point π/2 does not affect probability because a single point has probability zero in continuous distributions. Hence the probability equals 1. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Completely incorrect because almost all points belong to the branch.
Option B → No halving occurs in the interval measure.
Option C → Probability values cannot equal π/2.

🎯 Strategy

Used: Elimination

Application:

Use continuous probability properties and principal branch definition.

Final Logic:

Removing one point from a continuous interval keeps probability equal to 1.

🧠 Memory Trick

"Single point ⇒ zero probability."

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

12 Which statement represents an incorrect step regarding the simplification of inverse trigonometric transformations?

A tan⁻¹(tan x) simplifies perfectly to x for all unrestricted real values x ∈ R.

B To effectively simplify inverse expressions involving √(1-x²), one substitutes x = sin θ or cos θ.

C The form tan⁻¹((cosx-sinx)/(cosx+sinx)) correctly simplifies to π/4 - x under bound x.

D 3sin⁻¹x formally yields sin⁻¹(3x - 4x³) mapped safely inside [-1/2, 1/2].

💡 Short Explanation

tan⁻¹(tanx)=x only in principal branch. Tangent is periodic. Outside branch, adjustment is needed.

✅ Correct Answer Explanation

The identity: tan⁻¹(tanx)=x is valid only for: x ∈ (−π/2, π/2). Because tangent is periodic, values outside the principal branch require correction. Hence option A is incorrect. Options B, C, and D are standard simplification methods or valid restricted identities from NCERT.

❓ Why Other Options Are Incorrect

Option B → Standard substitution used in inverse trigonometric simplification.
Option C → Derived from tangent subtraction identity correctly.
Option D → Valid under required restricted domain conditions.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "all unrestricted real values" signals overgeneralization.

Final Logic:

Inverse tangent identities hold only within principal branch intervals.

🧠 Memory Trick

"tan inverse works only inside branch."

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

📄 Passage

We know from Chapter 1, that if y = f(x) is an invertible function, then x = f⁻¹(y). Thus, the graph of sin⁻¹ function can be obtained from the graph of original function by interchanging x and y axes, i.e., if (a, b) is a point on the graph of sine function, then (b, a) becomes the corresponding point on the graph of inverse of sine function. The dark portion of the graph of y = sin⁻¹ x represent the principal value branch.

Based directly on the passage, what mapping process safely obtains the graph of y = sin⁻¹x from y = sin x?

A By shifting the graph's continuous region by π/2.

B By systematically interchanging the mapping of x and y axes.

C By mathematically inverting the output function values.

D By multiplying the inputs explicitly by -1.

💡 Short Explanation

Inverse graphs reflect across y=x. Coordinates interchange positions. (a,b) becomes (b,a).

✅ Correct Answer Explanation

For any invertible function: if y=f(x), then x=f⁻¹(y). Graphically, the inverse function is obtained by interchanging x and y coordinates. Thus every point: (a,b) on y=sinx becomes: (b,a) on y=sin⁻¹x. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → No translation by π/2 occurs.
Option C → Inverse functions do not invert outputs algebraically.
Option D → Multiplying by −1 gives reflection, not inversion.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Directly identify the graphical rule stated in the passage.

Final Logic:

Inverse graphs are obtained by swapping x and y coordinates.

🧠 Memory Trick

"Inverse means swap coordinates."

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

📄 Passage

If the coordinate point (π/6, 1/2) visibly lies on the graph of the sine function, which corresponding geometric point must sit on the principal value branch of y = sin⁻¹x?

A (-π/6, -1/2)

B (1/2, -π/6)

C (1/2, π/6)

D (π/6, 2)

💡 Short Explanation

Inverse graph swaps coordinates. (a,b) becomes (b,a). Thus point becomes (1/2, π/6).

✅ Correct Answer Explanation

The point: (π/6,1/2) lies on y=sinx because: sin(π/6)=1/2. For inverse graphs, coordinates interchange. Hence corresponding point on: y=sin⁻¹x becomes: (1/2,π/6). Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Reflects through origin incorrectly.
Option B → Wrong sign for angle value.
Option D → Coordinates not interchanged properly.

🎯 Strategy

Used: Substitution

Application:

Apply coordinate interchange rule directly.

Final Logic:

Inverse mapping swaps x and y positions.

🧠 Memory Trick

"Swap to invert."

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

15 Evaluate and simplify the algebraic form tan⁻¹(x/√(a²-x²)) logically where |x| < a.

A sin⁻¹(a/x)

B cos⁻¹(x/a)

C tan⁻¹(a/x)

D sin⁻¹(x/a)

💡 Short Explanation

Put x=a sinθ. Expression reduces to tanθ. Result becomes θ=sin⁻¹(x/a).

✅ Correct Answer Explanation

Let: x=a sinθ, where: |x|<a. Then: √(a²−x²)=a cosθ. Thus: tan⁻¹(x/√(a²−x²)) = tan⁻¹(tanθ) = θ within principal branch. Since: θ=sin⁻¹(x/a), the simplified form is: sin⁻¹(x/a).

❓ Why Other Options Are Incorrect

Option A → Reciprocal expression invalid.
Option B → Corresponds to complementary angle.
Option C → Does not simplify the radical form.

🎯 Strategy

Used: Substitution

Application:

Use trigonometric substitution x=a sinθ.

Final Logic:

Expression reduces directly to tanθ.

🧠 Memory Trick

"x=a sinθ removes root."

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

16 Break down the rational expression tan⁻¹[(3a²x - x³) / (a³ - 3ax²)] into its most simplified identical format.

A 3 tan⁻¹(x/a)

B 3 sin⁻¹(x/a)

C a tan⁻¹x

D 1/3 tan⁻¹(x/a)

💡 Short Explanation

Use tan3θ identity. Put tanθ=x/a. Expression becomes tan⁻¹(tan3θ).

✅ Correct Answer Explanation

Using: tan3θ = (3tanθ−tan³θ)/(1−3tan²θ). Let: tanθ=x/a. Then the expression becomes: tan⁻¹(tan3θ)=3θ inside principal restrictions. Therefore: 3θ=3tan⁻¹(x/a). Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Uses sine inverse incorrectly.
Option C → Arbitrary scaling factor introduced.
Option D → Triple-angle identity reversed incorrectly.

🎯 Strategy

Used: Substitution

Application:

Recognize triple-angle tangent structure.

Final Logic:

Expression matches tan3θ identity exactly.

🧠 Memory Trick

"3a²x−x³ ⇒ tan triple-angle."

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

17 Deduce the numerical evaluation of the identity proof fragment: cos⁻¹(1/2) + 2sin⁻¹(1/2).

A π/3

B 2π/3

C π

D 3π/4

💡 Short Explanation

cos⁻¹(1/2)=π/3. sin⁻¹(1/2)=π/6. Total becomes 2π/3.

✅ Correct Answer Explanation

Using principal values: cos⁻¹(1/2)=π/3 and: sin⁻¹(1/2)=π/6. Hence: π/3 + 2(π/6) = π/3 + π/3 = 2π/3. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores second term.
Option C → Incorrect total addition.
Option D → Uses wrong inverse values.

🎯 Strategy

Used: Substitution

Application:

Replace inverse expressions with standard exact angles.

Final Logic:

π/3 + π/3 = 2π/3.

🧠 Memory Trick

"½ gives 30° and 60°."

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

18 Using domain conditions, precisely map the continuous range applicable for sec⁻¹x.

A [-π/2, π/2]

B (0, π)

C [0, π] - {π/2}

D [-1, 1]

💡 Short Explanation

sec inverse excludes π/2. Principal branch spans 0 to π. sec undefined at π/2.

✅ Correct Answer Explanation

The principal range of sec⁻¹x is: [0,π]−{π/2}. The point π/2 is excluded because: sec(π/2) is undefined. Thus option C correctly describes the principal range. Other options correspond to different inverse trigonometric functions or incomplete intervals.

❓ Why Other Options Are Incorrect

Option A → Belongs to sine inverse related range.
Option B → Excludes valid endpoints 0 and π.
Option D → Represents domain of sin inverse.

🎯 Strategy

Used: Elimination

Application:

Recall standard principal range definitions from NCERT.

Final Logic:

sec inverse range excludes only π/2.

🧠 Memory Trick

"sec skips π/2."

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

19 Execute the evaluation of tan⁻¹(√3) - sec⁻¹(-2) drawn strictly from Exercise 2.1 practice.

A π

B -π

C π/3

D -π/3

💡 Short Explanation

tan⁻¹(√3)=π/3. sec⁻¹(−2)=2π/3. Difference equals −π/3.

✅ Correct Answer Explanation

Using principal values: tan⁻¹(√3)=π/3. Also: sec⁻¹(−2)=2π/3 because: sec(2π/3)=−2. Therefore: π/3 − 2π/3 = −π/3. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Adds instead of subtracting.
Option B → Magnitude incorrectly doubled.
Option C → Ignores second term.

🎯 Strategy

Used: Substitution

Application:

Replace inverse trigonometric expressions with exact principal angles.

Final Logic:

π/3−2π/3=−π/3.

🧠 Memory Trick

"√3 → 60°, −2 sec → 120°."

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

20 Evaluate the nested trigonometric combination formula tan⁻¹[ 2cos(2sin⁻¹(1/2)) ].

A π/4

B π/2

C π/3

D π/6

💡 Short Explanation

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

✅ Correct Answer Explanation

First: sin⁻¹(1/2)=π/6. Then: 2sin⁻¹(1/2)=π/3. Now: cos(π/3)=1/2. Hence: 2cos(π/3)=1. Finally: tan⁻¹(1)=π/4. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → tan⁻¹(1) is not π/2.
Option C → Confuses cosine value with final inverse tangent.
Option D → Uses incorrect final evaluation.

🎯 Strategy

Used: Substitution

Application:

Evaluate nested inverse and trigonometric expressions stepwise.

Final Logic:

tan⁻¹(1)=π/4.

🧠 Memory Trick

"Half → π/6 → cosπ/3 → 1."

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

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