Principal Values MCQ Quiz in தமிழ் - Objective Question with Answer for Principal Values - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Apr 2, 2025
Latest Principal Values MCQ Objective Questions
Top Principal Values MCQ Objective Questions
Principal Values Question 1:
Find the value of sin-1 (0.5) ?
Answer (Detailed Solution Below)
Principal Values Question 1 Detailed Solution
Concept:
| 0° | 30° | 45° | 60° | 90° | |
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | ∞ |
| csc | ∞ | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | ∞ |
| cot | ∞ | √3 | 1 | 1/√3 | 0 |
Calculation:
We have to find the value of sin-1 (0.5)
Let sin-1 (0.5) = θ
⇒ sin θ = 0.5 =
⇒ sin θ = sin 30°
∴ θ = 30°
Hence θ = sin-1 (0.5) = 30°
Principal Values Question 2:
Find the principal value of
Answer (Detailed Solution Below)
Principal Values Question 2 Detailed Solution
Concept:
- cot-1 (-x) = π - cot-1x
-
cot (π/3) = 1/√3
Calculation:
Let y =
⇒ y = π - π/3
⇒ y = 2π/3
Since range of cot-1 is (0, π)
Hence, the principal value is
Principal Values Question 3:
Find the principal value of cos-1 (√3/2) + cos-1 (-1/2)?
Answer (Detailed Solution Below)
Principal Values Question 3 Detailed Solution
Concept:
cos-1 (-x) = π - cos-1 x, x ∈ [-1, 1]
cos-1 (cos θ) = θ, ∀ θ ∈ [0, π]
Calculation:
As we know that, cos-1 (-x) = π - cos-1 x, x ∈ [-1, 1]
⇒ cos-1 (-1/2) = π - cos-1 (1/2)
⇒ cos-1 (√3/2) + cos-1 (-1/2) = cos-1 (√3/2) + [π - cos-1 (1/2)]
As we know that, cos π/3 = 1/2 and cos π/6 = √3/2
⇒ cos-1 (√3/2) + cos-1 (-1/2) = cos-1 (cos π/6) + [π - cos-1 (cos π/3)]
As we know that, cos-1 (cos θ) = θ, ∀ θ ∈ [0, π]
⇒ cos-1 (√3/2) + cos-1 (-1/2) = π/6 + [π - π/3] = 5π/6
Principal Values Question 4:
Evaluate:
Answer (Detailed Solution Below)
Principal Values Question 4 Detailed Solution
Concept:
| sin-1 (-x) | - sin-1 x | cos-1 (-x) | π - cos-1 x |
| cosec-1 (-x) | - cosec-1 x | sec-1 (-x) | π - sec-1 x |
| tan-1 (-x) | - tan-1 x | cot-1 (-x) | π - cot-1 x |
tan (- x) = - tan x
cos (- x) = cos x
sin (- x) = - sin x
tan (2π - x) = - tan x
Calculation:
Given:
tan-1 (- 1) = - tan-1 (1) =
Additional Information
| Function | Domain | Range of Principal Value |
| sin-1 x | [-1, 1] | [-π/2, π/2] |
| cos-1 x | [-1, 1] | [0, π] |
| csc-1 x | R - (-1, 1) | [-π/2, π/2] - {0} |
| sec-1 x | R - (-1, 1) | [0, π] - {π/2} |
| tan-1 x | R | (-π/2, π/2) |
| cot-1 x | R | (0, π) |
Principal Values Question 5:
The value of
Answer (Detailed Solution Below)
Principal Values Question 5 Detailed Solution
Explanation:
If sinθ = x ⇒ θ = sin-1x, for θ ∈ [-π/2, π/2]
cot (cot-1 x) =x for x ∈ R
We have,
Let
⇒ cosθ = 7/25
⇒ sinθ = 24/25
⇒ cotθ = 7/24
∴
= cotθ
= 7/24
Principal Values Question 6:
Comprehension:
Directions: Read the following information carefully and answer the questions given below.
If
is equal to
Answer (Detailed Solution Below)
Principal Values Question 6 Detailed Solution
Concept:
Trigonometric ratios:
- sin θ =
- cos θ =
- tan θ =
Trigonometric Formulae:
- cos 3θ = 4 cos3θ − 3 cos θ
- cos(π − θ) = − cos θ
Calculation:
In anc Δ ABC and Δ PQR, by using trigonometry ratio formula,
cos α =
Now we know that
cos 3α = 4 cos3α − 3 cos α
⇒ cos 3α =
⇒ cos 3α =
⇒ cos 3α =
Since, cos(π − θ) = − cos θ
⇒ cos(π − 3α) = − cos 3α
⇒ cos(π − 3α) =
⇒ (π − 3α) = cos-1
∴ The value of cos-1(
Principal Values Question 7:
Find the principal value of
Answer (Detailed Solution Below)
Principal Values Question 7 Detailed Solution
Concept:
| Function | Domain | Range of Principal Value |
| sin-1 x | [-1, 1] | [-π/2, π/2] |
| cos-1 x | [-1, 1] | [0, π] |
| cosec-1 x | R - (-1, 1) | [-π/2, π/2] - {0} |
| sec-1 x | R - (-1, 1) | [0, π] - {π/2} |
| tan-1 x | R | (-π/2, π/2) |
| cot-1 x | R | (0, π) |
Inverse Trigonometric Functions for Negative Arguments:
| sin-1 (-x) | - sin-1 x | cos-1 (-x) | π - cos-1 x |
| cosec-1 (-x) | - cosec-1 x | sec-1 (-x) | π - sec-1 x |
| tan-1 (-x) | - tan-1 x | cot-1 (-x) | π - cot-1 x |
Calculation:
As we know sin-1 (-x) = - sin-1 x
So,
Let
⇒ sin θ =
∴ θ = 30°
Hence,
Principal Values Question 8:
Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying
Answer (Detailed Solution Below)
Principal Values Question 8 Detailed Solution
Calculation
Given
tan-1x + tan-1 2x =
⇒ tan-1 2x =
Taking tan both sides
⇒ 2x =
⇒ 2x2 + 3x - 1 = 0
⇒ x =
Only possible x =
Hence option (2) is correct
Principal Values Question 9:
The principal value of
Answer (Detailed Solution Below)
Principal Values Question 9 Detailed Solution
We have
Principal Values Question 10:
The value of
Answer (Detailed Solution Below)
Principal Values Question 10 Detailed Solution
Given:
Formula Used:
2cos-1x = cos-1(2x2 - 1)
Calculation:
We have,
⇒
⇒
⇒
⇒
⇒
⇒
⇒
∴ The value of