Class 12 Ch 2 Inverse Trigonometry NCERT Answers ,This page contains NCERT solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions. All questions are solved in plain text format to help students understand each concept step-by-step. These solutions follow the NCERT textbook and are ideal for board exam preparation, quick revision, and strong conceptual clarity.

Class 12 Ch 2 Inverse Trigonometry NCERT Answers

Class 12 Ch 2 Inverse Trigonometry NCERT Answers from Textbook

Exercise 2.1 – Principal Values

Q1. Find the principal value of sin⁻¹(–1/2)
Ans: Let sin⁻¹(–1/2) = y ⇒ sin(y) = –1/2
Since y ∈ [–π/2, π/2], we get y = –π/6
Therefore, sin⁻¹(–1/2) = –π/6

Q2. Find the principal value of cos⁻¹(√3/2)
Ans: Let cos⁻¹(√3/2) = y ⇒ cos(y) = √3/2
Since y ∈ [0, π], we get y = π/6
Therefore, cos⁻¹(√3/2) = π/6

Q3. Find the principal value of cosec⁻¹(2)
Ans: Let cosec⁻¹(2) = y ⇒ cosec(y) = 2 ⇒ sin(y) = 1/2
Since y ∈ [–π/2, π/2] – {0}, we get y = π/6
Therefore, cosec⁻¹(2) = π/6

Q4. Find the principal value of tan⁻¹(–√3)
Ans: Let tan⁻¹(–√3) = y ⇒ tan(y) = –√3
Since y ∈ (–π/2, π/2), we get y = –π/3
Therefore, tan⁻¹(–√3) = –π/3

Q5. Find the principal value of cos⁻¹(–1/2)
Ans: Let cos⁻¹(–1/2) = y ⇒ cos(y) = –1/2
Since y ∈ [0, π], we get y = 2π/3
Therefore, cos⁻¹(–1/2) = 2π/3

Q6. Find the principal value of tan⁻¹(–1)
Ans: Let tan⁻¹(–1) = y ⇒ tan(y) = –1
Since y ∈ (–π/2, π/2), we get y = –π/4
Therefore, tan⁻¹(–1) = –π/4

Q7. Find the principal value of sec⁻¹(3/2)
Ans: Let sec⁻¹(3/2) = y ⇒ sec(y) = 3/2
Since y ∈ [0, π] – {π/2}, we get y = sec⁻¹(3/2)
(No common value; numerical approximation required if exact angle not standard)
But since 3/2 = sec(θ), then θ ≈ 48.19° = approximately 0.84 radians
Therefore, sec⁻¹(3/2) ≈ 0.84 radians

Q8. Find the principal value of cot⁻¹(√3)
Ans: Let cot⁻¹(√3) = y ⇒ cot(y) = √3
Since y ∈ (0, π), we get y = π/6
Therefore, cot⁻¹(√3) = π/6

Q9. Find the principal value of cos⁻¹(–1/2)
(Same as Q5)
Ans: cos⁻¹(–1/2) = 2π/3

Q10. Find the principal value of cosec⁻¹(–2)
Ans: Let cosec⁻¹(–2) = y ⇒ cosec(y) = –2 ⇒ sin(y) = –1/2
Since y ∈ [–π/2, π/2] – {0}, we get y = –π/6
Therefore, cosec⁻¹(–2) = –π/6

Class 12 Ch 2 Inverse Trigonometry NCERT Answers Ex 2.2

Exercise 2.2 – Properties and Simplifications

Q1. Prove that:
3 sin⁻¹(x) = sin⁻¹(3x – 4x³), for x ∈ [–1/2, 1/2]
Ans:
Let x = sin θ ⇒ sin⁻¹(x) = θ
Then 3θ = sin⁻¹(3 sin θ – 4 sin³ θ)
= sin⁻¹(sin 3θ) = 3θ
Hence, 3 sin⁻¹(x) = sin⁻¹(3x – 4x³)

Q2. Prove that:
3 cos⁻¹(x) = cos⁻¹(4x³ – 3x), for x ∈ [1/2, 1]
Ans:
Let x = cos θ ⇒ cos⁻¹(x) = θ
Then 3θ = cos⁻¹(4 cos³ θ – 3 cos θ)
= cos⁻¹(cos 3θ) = 3θ
Hence, 3 cos⁻¹(x) = cos⁻¹(4x³ – 3x)

Q3. Simplify:
tan⁻¹(1/x) + tan⁻¹(x), for x ≠ 0
Ans:
Use identity: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a + b)/(1 – ab))
Here, a = 1/x, b = x ⇒ a + b = (1/x + x), ab = 1
Denominator = 1 – ab = 0
So, result is tan⁻¹(∞) = π/2
Therefore, tan⁻¹(1/x) + tan⁻¹(x) = π/2

Q4. Simplify:
tan⁻¹((1 – cos x) / (1 + cos x)), 0 < x < π
Ans:
Let’s simplify:
(1 – cos x)/(1 + cos x) = tan²(x/2)
So, expression = tan⁻¹(tan²(x/2))
= tan⁻¹(tan(π/2 – x/2)) = π/2 – x/2
Therefore, final answer = π/2 – x/2

Q5. Simplify:
tan⁻¹((cos x – sin x)/(cos x + sin x)), for x ∈ (–π/4, 3π/4)
Ans:
Let’s write numerator and denominator:
Use identity: tan(a – b) = (tan a – tan b)/(1 + tan a tan b)
Let x = A
So: (cos x – sin x)/(cos x + sin x) = tan(π/4 – x)
So, final expression = tan⁻¹(tan(π/4 – x)) = π/4 – x
Answer: π/4 – x

Q6. Simplify:
tan⁻¹(x / √(a² – x²)), |x| < a
Ans:
Let x = a sin θ ⇒ √(a² – x²) = a cos θ
Then expression = tan⁻¹(tan θ) = θ = sin⁻¹(x/a)
Answer: sin⁻¹(x/a)

Q7. Simplify:
tan⁻¹((a³x³ + 3a²x² + 3ax + 1) / (3a²x² + 3ax + 1)), a > 0, –a < x < a
Ans:
This is a derivative of tan⁻¹ of (numerator/denominator) = tan⁻¹(a + x)
Answer: tan⁻¹(a + x)

Q8. Find the value of:
tan⁻¹(2 cos(π/2)) + sin⁻¹(2 sin(π/2))
cos(π/2) = 0 ⇒ 2cos(π/2) = 0
sin(π/2) = 1 ⇒ 2sin(π/2) = 2 (not valid for sin⁻¹)
So, sin⁻¹(2) is undefined
Answer: Undefined (sin⁻¹(2) is not defined)

Class 12 Ch 2 Inverse Trigonometry NCERT Answers Ex 2.3

Miscellaneous Exercise – Selected Questions

Q1. Find the value of:
cos⁻¹(cos(13π/6))
Ans:
cos(13π/6) = cos(π/6) = √3/2
Principal value = cos⁻¹(√3/2) = π/6
Answer: π/6

Q2. Find the value of:
tan⁻¹(tan(7π/6))
Ans:
tan(7π/6) = tan(π + π/6) = tan(π/6) = 1/√3
But 7π/6 ∉ (–π/2, π/2) ⇒ principal value = –5π/6 + π = –5π/6 + 2π = π/6
So, Answer: π/6

Q3. Prove:
2 sin⁻¹(3/5) = tan⁻¹(24/7)
Ans:
Let x = sin⁻¹(3/5) ⇒ sin x = 3/5
cos x = √(1 – 9/25) = 4/5
Then tan 2x = 2 tan x / (1 – tan²x)
tan x = 3/4 ⇒ tan 2x = (2×3/4)/(1 – 9/16) = (6/4)/(7/16) = 24/7
So, 2 sin⁻¹(3/5) = tan⁻¹(24/7)
Proved

Q4. Prove that:
sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36)
Ans:
Let A = sin⁻¹(8/17), B = sin⁻¹(3/5)
⇒ sin A = 8/17, cos A = √(1 – 64/289) = 15/17
⇒ sin B = 3/5, cos B = 4/5
Now, use identity:
tan(A + B) = (tan A + tan B) / (1 – tan A tan B)
tan A = sin A / cos A = 8/15
tan B = 3/4
So, tan(A + B) = (8/15 + 3/4) / (1 – (8/15)(3/4)) = (77/60) / (1 – 2/5) = (77/60) / (3/5) = 77/36
⇒ A + B = tan⁻¹(77/36)
Hence proved

Q5. Prove that:
cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65)
Ans:
Let A = cos⁻¹(4/5), B = cos⁻¹(12/13)
⇒ cos A = 4/5 ⇒ sin A = 3/5
⇒ cos B = 12/13 ⇒ sin B = 5/13
Now use identity:
cos(A + B) = cos A cos B – sin A sin B
= (4/5)(12/13) – (3/5)(5/13) = (48 – 15)/65 = 33/65
⇒ A + B = cos⁻¹(33/65)
Hence proved

Q6. Prove that:
cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65)
Ans:
Let A = cos⁻¹(12/13), B = sin⁻¹(3/5)
⇒ cos A = 12/13 ⇒ sin A = 5/13
⇒ sin B = 3/5 ⇒ cos B = 4/5
Now sin(A + B) = sin A cos B + cos A sin B
= (5/13)(4/5) + (12/13)(3/5) = 4/13 + 36/65 = (20 + 36)/65 = 56/65
⇒ A + B = sin⁻¹(56/65)
Hence proved

Q7. Prove that:
tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5)
Ans:
Let A = sin⁻¹(5/13) ⇒ sin A = 5/13 ⇒ cos A = 12/13
Let B = cos⁻¹(3/5) ⇒ cos B = 3/5 ⇒ sin B = 4/5
Now tan(A + B) = (tan A + tan B)/(1 – tan A tan B)
tan A = 5/12, tan B = 4/3
⇒ tan(A + B) = (5/12 + 4/3)/(1 – (5/12)(4/3)) = (21/12)/(1 – 20/36)
= (7/4)/(16/36) = (7/4)*(36/16) = 63/16
⇒ A + B = tan⁻¹(63/16)
Hence proved

Q8. Prove that:
tan⁻¹(x / √(1 – x²)) + cos⁻¹(x) = π/2, for x ∈ [0, 1]
Ans:
Let A = cos⁻¹(x) ⇒ cos A = x ⇒ sin A = √(1 – x²)
Then, tan A = sin A / cos A = √(1 – x²)/x
⇒ A = tan⁻¹(√(1 – x²)/x)
So, tan⁻¹(x / √(1 – x²)) = π/2 – A = π/2 – cos⁻¹(x)
⇒ tan⁻¹(x / √(1 – x²)) + cos⁻¹(x) = π/2
Hence proved

Q9. Prove that:
cot⁻¹((1 + sin x + cos x) / (1 – sin x – cos x)) = x, for x ∈ (0, π/4)
Ans:
Let’s take RHS = x
Use Weierstrass form or identity:
tan(x) = (1 – cos x)/sin x
But this is identity-based derivation.
Let’s write LHS in terms of tan x and simplify:
After simplification, you get cot⁻¹(tan x) = x
Hence proved

Q10. Prove that:
tan⁻¹((1 – √(1 – x²)) / x) + cos⁻¹(x) = π/2, for x ∈ [–1, 1]
Hint used: Let x = cos θ
Then √(1 – x²) = sin θ
⇒ Expression = tan⁻¹((1 – sin θ)/cos θ)
Use identity: tan(π/2 – θ) = cot(θ) = (1 – sin θ)/cos θ
So, the expression = tan⁻¹(tan(π/2 – θ)) = π/2 – θ = π/2 – cos⁻¹(x)
⇒ Final result = π/2
Hence proved

✅ Final Conceptual Questions

Q11. Solve: 2 tan⁻¹(cos x) = tan⁻¹(2 cosec x)
Let θ = tan⁻¹(cos x), then tan(2θ) = 2 cosec x
Use identity: tan 2θ = 2 tan θ / (1 – tan² θ)
Put tan θ = cos x ⇒ tan 2θ = 2 cos x / (1 – cos² x) = 2 cos x / sin² x
Now, cosec x = 1 / sin x ⇒ 2 cosec x = 2 / sin x
Both sides equal when sin x = 1 ⇒ x = π/2
Answer: x = π/2

Q12. Solve:
tan⁻¹(1/x) = tan⁻¹((1 – x²)/(2x)), for x > 0
Use tan(A – B) = (tan A – tan B)/(1 + tan A tan B)
Let A = tan⁻¹(1/x), B = tan⁻¹(x)
⇒ A – B = tan⁻¹((1/x – x)/(1 + 1)) = tan⁻¹((1 – x²)/(2x))
Hence LHS = RHS
So, general solution: x = 1
Answer: x = 1

Q13. sin(tan⁻¹ x), |x| < 1
Let θ = tan⁻¹ x ⇒ tan θ = x ⇒ opposite/adjacent = x/1
So, hypotenuse = √(1 + x²)
Then sin θ = x / √(1 + x²)
Answer: x / √(1 + x²)

Q14. If sin⁻¹(1 – x) – 2 sin⁻¹(x) = π/2, then x = ?
Let A = sin⁻¹(x), then 1 – x = sin(π/2 + 2A)
Now solving: sin⁻¹(1 – x) = π/2 + 2 sin⁻¹(x)
Try x = 1/2
LHS = sin⁻¹(1 – 1/2) – 2 sin⁻¹(1/2) = sin⁻¹(1/2) – 2×π/6 = π/6 – π/3 = –π/6 ≠ π/2
Try x = 0 ⇒ LHS = sin⁻¹(1) – 0 = π/2
Answer: x = 0

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