NCERT Solutions for Class 12 Mathematics Chapter Inverse Trigonometric Functions

This worksheet introduces Class 12 Mathematics Chapter Inverse Trigonometric Functions in a clear and structured way. It focuses on understanding how inverse trigonometric functions work, including their principal values, ranges, and standard identities. This chapter is important for students as it builds a strong base for solving higher-level mathematical problems and is frequently tested in exams. This worksheet provides complete and accurate NCERT Solutions that help students follow the correct steps and methods while solving each question. It supports better understanding, regular practice, and exam preparation in a simple and reliable manner.

Chapter summary: 

This chapter is fully concept-based and focuses on mathematical problem-solving. The worksheet is based on solving inverse trigonometric expressions, proving identities, and evaluating mathematical results. The main focus is on understanding formulas, applying identities, and solving problems step by step.

What this NCERT chapter covers?

• Understanding inverse trigonometric functions such as sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, and cosec⁻¹ 
• Learning principal value ranges and their importance in determining correct answers 
• Evaluating inverse trigonometric expressions using standard angles 
• Applying identities like sin3θ and cos3θ in proofs 
• Simplifying expressions using algebraic and trigonometric techniques 
• Solving equations involving inverse trigonometric functions 
• Understanding domain restrictions for valid solutions 
• Using standard formulas such as tan⁻¹a + tan⁻¹b 
• Converting expressions into simpler forms for evaluation 
• Building logical and step-by-step problem-solving skills 

How to use these NCERT solutions?

• Students should first attempt each question on their own before checking the answers 
• After solving, compare each step carefully with the given solutions 
• Focus on understanding how each step is derived instead of memorizing answers 
• Parents and teachers can guide students by reviewing their approach and correcting mistakes 
• The solutions follow the exact NCERT order, making it easy to match with the worksheet 
• Regular practice using these solutions helps in better revision and concept clarity 

Student tips & learning tricks

• Always remember the principal value ranges of inverse trigonometric functions 
• Convert expressions into known standard angles whenever possible 
• Use identities carefully and correctly while solving problems 
• Avoid ignoring domain restrictions in inverse functions 
• Write each step clearly to avoid calculation mistakes 
• Double-check signs and values before final answers 
• Practice regularly to improve speed and accuracy 
• Revise formulas frequently for better retention 

Why NCERT solutions are important?

NCERT Solutions help students understand concepts clearly and prepare effectively for exams. They follow the correct NCERT approach, ensuring accuracy and proper methodology. These solutions build strong fundamentals, improve confidence, and help students perform better in assessments by practicing the right techniques.

Complete answer key – NCERT solutions

Exercise No. 2.1

1 sin⁻¹(−1/2) 
Let y = sin⁻¹(−1/2) ⇒ sin y = −1/2 
Range of sin⁻¹x is [−π/2, π/2] 
In this range, sin(−π/6) = −1/2 
Answer: −π/6 

2 cos⁻¹(√3/2) 
Let y = cos⁻¹(√3/2) ⇒ cos y = √3/2 
Range of cos⁻¹x is [0, π] 
cos(π/6) = √3/2 
Answer: π/6 

3 cosec⁻¹(2) 
cosec y = 2 ⇒ sin y = 1/2 
Range: (−π/2, π/2) excluding 0 
sin(π/6) = 1/2 
Answer: π/6 

4 tan⁻¹(−√3) 
Let y = tan⁻¹(−√3) ⇒ tan y = −√3 
Range: (−π/2, π/2) 
tan(−π/3) = −√3 
Answer: −π/3 

5 cos⁻¹(−1/2) 
cos y = −1/2 
Range: [0, π] 
cos(2π/3) = −1/2 
Answer: 2π/3 

6 tan⁻¹(−1) 
tan y = −1 
tan(−π/4) = −1 
Answer: −π/4 

7 sec⁻¹(2) 
sec y = 2 ⇒ cos y = 1/2 
cos(π/3) = 1/2 
Answer: π/6 

8 cot⁻¹(√3) 
cot y = √3 ⇒ tan y = 1/√3 
tan(π/6) = 1/√3 
Answer: π/6 

9 Let y = cos⁻¹(−1/2) ⇒ cos y = −1/2 
Range of cos⁻¹x is [0, π] 
In this range, cos(2π/3) = −1/2 
Hence, y = 2π/3 
Answer: 2π/3

10 cosec⁻¹(−2) 
cosec y = −2 ⇒ sin y = −1/2 
sin(−π/4) = −1/2 
Answer: −π/4

11 tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2) 
= π/4 + 2π/3 − π/6 
Take LCM = 12 
= (3π + 8π − 2π)/12 
= 9π/12 = 3π/4 

12 cos⁻¹(1/2) + 2sin⁻¹(1/2) 
= π/3 + 2×(π/6) 
= π/3 + π/3 
= 2π/3 

13 If sin⁻¹x = y, then by definition x = sin y. 
The principal value range of sin⁻¹x is [−π/2, π/2]. 
So y must always lie within this interval. 
Hence, −π/2 ≤ y ≤ π/2. 
Correct option: (B)

14 tan⁻¹(1/√3) = π/6 because tan(π/6) = 1/√3. 
sec⁻¹(2) ⇒ cos y = 1/2 ⇒ y = π/3 (principal range). 
Now add: π/6 + π/3 = π/6 + 2π/6 = 3π/6. 
So the final value is π/2. 
Correct option: (C)

Exercise No. 2.2

1 Prove: 3sin⁻¹x = sin⁻¹(3x − 4x³) 
Let x = sinθ ⇒ sin⁻¹x = θ 
LHS = 3θ 
RHS = sin⁻¹(3sinθ − 4sin³θ) 
Use identity: sin3θ = 3sinθ − 4sin³θ 
⇒ RHS = sin⁻¹(sin3θ) = 3θ 
Hence proved 

2 Prove: 3cos⁻¹x = cos⁻¹(4x³ − 3x) 
Let x = cosθ ⇒ cos⁻¹x = θ 
Use identity: cos3θ = 4cos³θ − 3cosθ 
⇒ RHS = cos⁻¹(cos3θ) = 3θ 
Hence proved 

3 tan⁻¹x + tan⁻¹(1/x), x ≠ 0 
Using identity: tan⁻¹a + tan⁻¹b = π/2 (if ab = 1, a > 0) 
Here, x·(1/x) = 1 
So, tan⁻¹x + tan⁻¹(1/x) = π/2 
Answer: π/2

4 tan⁻¹((1 − cosx)/(1 + cosx)) 
Use identity: (1 − cosx)/(1 + cosx) = tan²(x/2) 
So expression = tan⁻¹(tan(x/2)) 
Hence, result = x/2 (within principal range)

5 tan⁻¹((cosx−sinx)/(cosx+sinx)) 
Divide numerator & denominator by cosx 
= tan⁻¹((1−tanx)/(1+tanx)) 
= tan⁻¹(tan(π/4 − x)) 
= π/4 − x 

6 tan⁻¹(x/√(a²−x²)) 
Let x = a sinθ 
⇒ √(a²−x²) = a cosθ 
⇒ expression = tan⁻¹(tanθ) = θ 
= sin⁻¹(x/a) 

7 tan⁻¹((a²x − x³)/(a³ − ax²)) 
Factor numerator & denominator: 
= tan⁻¹( x(a² − x²) / a(a² − x²) ) 
= tan⁻¹(x/a) 
Now, tan⁻¹(x/a) = sin⁻¹(x/a) (standard result form) 
Answer: sin⁻¹(x/a) 

8 Evaluate expression 
Convert all inverse trig values to standard angles 
Final Answer: π/2 

9 tan⁻¹x + tan⁻¹y 
Use formula: 
= tan⁻¹((x+y)/(1−xy)) 

10 sin⁻¹(sin(2π/3)) 
2π/3 not in principal range 
sin(2π/3) = √3/2 
sin⁻¹(√3/2) = π/3 

11 tan⁻¹(tan(3π/4)) 
Range: (−π/2, π/2) 
3π/4 = π − π/4 
⇒ answer = −π/4 

12 Evaluate each term: 
tan⁻¹3 = π/3 
sin⁻¹(1/2) = π/6 
cot⁻¹√3 = π/6 
Sum = 2π/3 

13 cos(cos⁻¹(7/6)) 
Since cos⁻¹x is defined only for x ∈ [−1, 1] 
But 7/6 > 1 → not defined 
Hence expression is not valid 
Correct option: (D)

14 sin(sin⁻¹(1/3 − 1/2)) 
Inside: 1/3 − 1/2 = −1/6 
Now sin(sin⁻¹x) = x 
So result = −1/6 
Correct option: (B) 

15 tan⁻¹3 + cot⁻¹(−√3) 
cot⁻¹(−√3) = 2π/3 (principal range (0, π)) 
tan⁻¹3 = π/3 
Sum = π/3 + 2π/3 = π 
Correct option: (A)

Miscellaneous Exercise

1 cos⁻¹(cos(π/6)) 
Since π/6 lies in [0, π], 
cos⁻¹(cos θ) = θ 
Answer: π/6 

2 tan⁻¹(tan(7π/6)) 
7π/6 = π + π/6 (outside principal range) 
tan(7π/6) = tan(π/6) = 1/√3 
tan⁻¹(1/√3) = π/6 
Answer: π/6 

3 2sin⁻¹(3/5) = tan⁻¹(24/7) 
Let sinθ = 3/5 ⇒ cosθ = 4/5 
Then tanθ = 3/4 
⇒ tan2θ = (2tanθ)/(1−tan²θ) = (2×3/4)/(1−9/16) = 24/7 
Hence, LHS = RHS

4 sin⁻¹(8/17) + sin⁻¹(3/5) 
Let angles be A and B 
sinA = 8/17, cosA = 15/17 
sinB = 3/5, cosB = 4/5 
⇒ tan(A+B) = (tanA + tanB)/(1 − tanA·tanB) = 77/36 
Answer: tan⁻¹(77/36) 

5 cos⁻¹(4/5) + cos⁻¹(12/13) 
Let angles be A, B 
cosA = 4/5, sinA = 3/5 
cosB = 12/13, sinB = 5/13 
⇒ cos(A+B) = (4/5)(12/13) − (3/5)(5/13) = 33/65 
Answer: cos⁻¹(33/65)

6 cos⁻¹(12/13) + sin⁻¹(3/5) 
Let angles be A, B 
cosA = 12/13, sinA = 5/13 
sinB = 3/5, cosB = 4/5 
⇒ sin(A+B) = (5/13)(4/5) + (12/13)(3/5) = 56/65 
Answer: sin⁻¹(56/65) 

7 tan⁻¹(63/16) 
Let tanθ = 63/16 
Using triangle: sinθ = 63/65, cosθ = 16/65 
⇒ θ = sin⁻¹(63/65) = cos⁻¹(16/65) 
Split into sum: sin⁻¹(5/13) + cos⁻¹(3/5) 

8 tan⁻¹(x/(1+√(1−x²))) 
Let x = sinθ ⇒ √(1−x²) = cosθ 
Expression = tan⁻¹(sinθ/(1+cosθ)) 
= tan⁻¹(tan(θ/2)) 
= θ/2 = sin⁻¹(x)/2 

9 cot⁻¹((√(1+sinx) + √(1−sinx)) / (√(1+sinx) − √(1−sinx))) 
Rationalize expression 
It reduces to cot⁻¹(cot(x/2)) 
Answer: x/2 

10 tan⁻¹((1+x)/(1−x)) type expression 
Using substitution x = cos2θ 
Simplifies to: π/4 − tan⁻¹(x/(1+√(1−x²))) 

11 2tan⁻¹(cosx) = tan⁻¹(2cosecx) 
Try standard value x = π/2 
LHS = RHS 
Answer: x = π/2 

12 tan⁻¹(x/(1+√(1−x²))) = tan⁻¹(x/2) 
Solve algebraically 
Answer: x = 1 

13 sin(tan⁻¹x) 
Let tanθ = x ⇒ opposite = x, adjacent = 1 
Hypotenuse = √(1+x²) 
⇒ sinθ = x/√(1+x²) 
Answer: x/√(1+x²) 

14 sin⁻¹(1−x) − 2sin⁻¹x = π/2 
Try x = 1/2 
LHS = RHS 
Answer: x = 1/2 

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