Question

Class 12 Maths MCQ Questions of Inverse Trigonometric Functions with Answers?

Answer 1

The free Class 12 MCQ Questions of Inverse Trigonometric Functions with Answers are ready by Sarthaks eConnect specialists as per the NCERT educational program. The Inverse Trigonometric Functions Class 12 Important MCQ Questions Covers Major Topics on Basic Concepts and Properties of Inverse trigonometric functions. These MCQ Questions are prepared so students get a bit-by-bit approach to solving any problems.

Inverse trigonometry is quite possibly the main point in the NCERT educational program for Class 12 students. The free MCQ Questions for Class 12 Maths additionally contain answers that are created as practices for students. So, they can improve their subject knowledge. Students can get to the free Multiple-Choice Questions accessible on Sarthaks to get ready for their tests.

Practice MCQ Question for Class 12 Maths

1. The value of cot (sin^–1 x) is

(a) √(1+x²) /x
(b) x/√(1+x²)
(c) 1/x
(d) √(1 – x²) /x

2. If tan-1 √(1 + x²)-1 /x = 4^∘, then

(a) x= tan 2°
(b) x = 4°
(c) x =tan(1/4)°
(d) x = tan 8°

3. What will be the value of x + y + z if cos^-1 x + cos^-1 y + cos^-1 z = 3π?
(a) -1/3
(b) 1
(c) 3
(d) -3

4. Which value is similar to sin^-1sin(6 π/7)?

(a) sin^-1(π/7)
(b) cos^-1(π/7)
(c) sin^-1(2π/7)
(d) coses^-1(π/7)

5. sec^-1 x + cosec^-1 x = ……………….

(a) π/2
(b) 2π/2
(c) π/4
(d) π/3

6. tan^-1 x + cot^-1 x = …………

(a) π/3
(b) 2π/2
(c) π/4
(d) π/2

7. Principal value of sin^-1 (-1/2) is ………………

(a) –π/6
(b) –π/5
(c) –π/2
(d) –5π/3

8. Principal value of sin^-1 (-1√2) is ………………

(a) –π/6
(b) –π/4
(c) –π/2
(d) –5π/3

9. What is the value of sin^-1(-x) for all x belongs to [-1, 1]?

(a) -sin^-1(x)
(b) sin^-1(x)
(c) 2sin^-1(x)
(d) sin^-1(-x)/2

10. Principal value of cos^-1 (−1/2) is ………………

(a) 2π/3
(b) π/4
(c) 3π/2
(d) 5π/3

11. Principal value of tan^-1 (-√3) is …………

(a) –π/6
(b) –π/5
(c) –π/3
(d) –5π/3

12. sin^-1 x + cos^-1 x = ………………

(a) π/4
(b) 2π/2
(c) π/2
(d) π/3

13. What is the value of sin^-1(sin 6)?

(a) -2π – 6
(b) 2π + 6
(c) -2π + 6
(d) 2π – 6

14. What is the value of cos^-1(-x) for all x belongs to [-1, 1]?

(a) cos^-1(-x)
(b) π – cos^-1(x)
(c) π – cos^-1(-x)
(d) π + cos^-1(x)

15. sin^-1(sin π/6)......

(a) π/2
(b) 2π/2
(c) π/6
(d) π/3

16. Principal values of cot^-1(-√3)

(a) π/2
(b) 3π/2
(c) π/6
(d) π/3

17. The value of tan(cos^-1(3/5) + tan^-1(1/4)) is 

(a) 19/8 
(b) 8/19 
(c) 19/12 
(d) 3/4

18. Number of solutions of the equation
tan^-1 (1/2x+1) + tan^-1 (1/4x+1) = tan^-1 (2/x²)

(a) 1
(b) 2
(c) 3
(d) 4

19. If 6 sin^-1 (x² – 6x + 8.5) = π, then x is equal to

(a) 1
(b) 2
(c) 3
(d) 8

20. The number of real solution of the equation is
√(1+cos2x)= √2 cos^-1(cos x) in [π/2, π] is

(a) 0
(b) 1
(c) 2
(d) None of these

Comments

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Answer 2

1. Answer: (d) √(1 – x²) /x

Explanation: Let sin⁻¹ x=θ , 

Then sinθ=x

⇒ cscθ = 1/x

⇒ csc² θ = 1/x²

⇒1+ cot² θ = 1/x²

⇒ cot = √(1 – x²) /x

2. Answer: (d) x = tan 8°

Explanation: Substituting x=tanA

tan^-1(sec A-1)/tanA = tan^-1(1-cosA)/sinA

= tan^-1(tan(A/2)).... Using cos A = 1-2sin² A/2 

4°=A/2

tan ⁻¹(x)=8°
⇒ A= 8°

x =tan (8°)

3. Answer: (d) -3

Explanation: The equation is cos^-1 x + cos^-1 y + cos^-1 z = 3π
This means cos^-1 x = π, cos^-1 y = π and cos^-1 z = π
This will be only possible when it is in maxima.
As, cos^-1 x = π so, x = cos^-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.

4. Answer: (a) sin^-1(π/7)

Explanation: sin^-1sin(6 π/7)
Now, sin(6 π/7) = sin(π – 6 π/7)
= sin(2π + 6 π/7) = sin(π/7)
= sin(3π – 6 π/7) = sin(20π/7)
= sin(-π – 6 π/7) = sin(-15π/7)
= sin(-2π + 6 π/7) = sin(-8π/7)
= sin(-3π – 6 π/7) = sin(-27π/7)
Therefore, sin^-1sin(6 π/7) = sin^-1(π/7).

5. Answer: (a) π/2

6. Answer: (d) π/2

7. Answer: (a) –π/6

8. Answer: (b) –π/4

9. Answer: (a) -sin^-1(x)

Explanation: Let, θ = sin^-1(-x)
So, -π/2 ≤ θ ≤ π/2
=> -x = sinθ
=> x = -sinθ
=> x = sin(-θ)
Also, -π/2 ≤ -θ ≤ π/2
=> -θ = sin^-1(x)
=> θ = -sin^-1(x)
So, sin^-1(-x) = -sin^-1(x)

10. Answer: (a) 2π/3

11. Answer: (c) –π/3

12. Answer: (c) π/2

13. Answer: (c) -2π + 6

Explanation: We know, sin x = sin(π – x)
So, sin 6 = sin(π – 6)
= sin(2π + 6)
= sin(3π – 6)
= sin(-π – 6)
= sin(-2π – 6)
= sin(-3π – 6)
So, sin^-1(sin 6) = sin^-1(sin (-2π + 6))
= -2π + 6

14. Answer: b

Explanation: Let, θ = cos^-1(-x)
So, 0 ≤ θ ≤ π
=> -x = cosθ
=> x = -cosθ
=> x = cos(-θ)
Also, -π ≤ -θ ≤ 0
So, 0 ≤ π -θ ≤ π
=> -θ = cos^-1(x)
=> θ = -cos^-1(x)
So, cos^-1(x) = π – θ
θ = π – cos^-1(x)
=> cos^-1(-x) = π – cos^-1(x)

15. Answer: (c) π/6

Explanation: Given sin^-1(sin π/6)

We know that the value of sin π/6 is 1/2

By substituting this value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Now let y = sin^-1 (1/2)

Sin (π/6) = 1/2

The range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = 1/2

Therefore sin^-1(sin π/6) = π/6

16. Answer: (c) π/6

Explanation: Given cot^-1(-√3)

Let y = cot^-1(-√3)

– Cot (π/6) = √3

= Cot (π – π/6)

= cot (5π/6)

Thus, the principal value of cot^-1 (- √3) is 5π/6

17. Answer: (a) 19/8 

Explanation: tan(cos^-13/5+tan^-11/4)=(tan^-14/3+tan^-11/4)

= tan tan^-1((4/3+1/4)/(1-4/3 x 1/4))

= tan tan^-1(19/8) =19/8

18. Answer: (b) 2

19. Answer: (b) 2

20. Answer: (c) 2

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