Inverse Trigonometric Functions MCQ Quiz - Objective Question with Answer for Inverse Trigonometric Functions - Download Free PDF
Last updated on Apr 22, 2025
Latest Inverse Trigonometric Functions MCQ Objective Questions
Inverse Trigonometric Functions Question 1:
Find value of cot (tan-1 x + cot-1 x)
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 1 Detailed Solution
Concept:
tan-1 x + cot-1 x =
Calculation:
As we know tan-1 x + cot-1 x =
∴ cot (tan-1 x + cot-1 x) = cot
Inverse Trigonometric Functions Question 2:
is
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 2 Detailed Solution
Concept:
The maxima and minima of any function f(x) can be calculated by keeping it's first derivative as zero.
Now to find out whether it is a maxima or minima, we take it's double derivative now if the value is negative then it is a maxima and if it is positive it is a minima. For value equals to zero, it is unstable.
Solution: Given function,
To find out whether it is a maxima or minima, we take the double derivative.
Hence, 4>0. It is minima at x = 90°.
The correct answer is option 1.
Inverse Trigonometric Functions Question 3:
The principal value of
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 3 Detailed Solution
Explanation:
If sinθ = x ⇒ θ = sin-1x, for θ ∈ [-π/2, π/2]
sin (sin-1 x) =x for -π/2 ≤ x ≤ π/2
We have,
= sin-1(sin(
∴ sin-1(sin(
Additional InformationPrincipal Values of Inverse Trigonometric Functions:
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) |
Inverse Trigonometric Functions Question 4:
If (tan(cos-1x) = sin(cot-1
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 4 Detailed Solution
Calculation:
Let, cot-1
⇒ cot ϕ =
⇒ sin ϕ =
Let cos-1x = θ
⇒ sec θ =
⇒ tan θ =
⇒ tan θ =
⇒ tan θ =
Now, (tan(cos-1x) = sin(cot-1
⇒ (tan(tan-1
⇒
⇒
Squaring both sides, we get:
(1 - x2)5 = 4x2
⇒ 9x2 = 5
⇒ x =
∴ The value of x is
The correct answer is Option 2.
Inverse Trigonometric Functions Question 5:
Principal value of
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 5 Detailed Solution
Explanation:
=
=
=
=
=
Option (2) is true.
Top Inverse Trigonometric Functions MCQ Objective Questions
If 4 tan-1 x + cot‑1 x = π, then x equals:
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 6 Detailed Solution
Download Solution PDFConcept:
Calculation:
4 tan-1 x + cot‑1 x = π
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 7 Detailed Solution
Download Solution PDFConcept:
Calculation:
S =
S =
S =
S =
S =
S =
S =
S =
The domain of sin-1 4x is:
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 8 Detailed Solution
Download Solution PDFConcept:
- The domain of a function f(x) is the set of values of x for which the function is defined.
- The value of sin θ always lies in the interval [-1, 1].
- sin-1 (sin θ) = θ.
- sin (sin-1 x) = x.
Calculation:
Let's say that sin-1 4x = θ.
⇒ sin (sin-1 4x) = sin θ
⇒ sin θ = 4x
Since, -1 ≤ sin θ ≤ 1
⇒ -1 ≤ 4x ≤ 1
⇒
⇒ x ∈
∴ The domain of the function is the closed interval
If sin-1 x + sin-1 y =
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 9 Detailed Solution
Download Solution PDFConcept:
sin-1 x + cos-1 x =
Calculation:
sin-1 x + sin-1 y =
⇒
⇒ π - ( cos-1 x + cos-1 y ) =
⇒ cos-1 x + cos-1 y =
The correct option is 2.
Find the value of
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 10 Detailed Solution
Download Solution PDFConcept:
Solution:
So, use the relation,
So,
If 3 sin-1 x + cos-1 x = π, then find the value of x?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 11 Detailed Solution
Download Solution PDFConcept:
sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]
Calculation:
Given: 3 sin-1 x + cos-1 x = π
⇒ 3 sin-1 x + cos-1 x = 2 sin-1 x + [sin-1 x + cos-1 x] = π
As we know that, sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]
⇒ 2 sin-1 x + [π /2] = π
⇒ 2 sin-1 x = π - π/2
⇒ 2 sin-1 x = π/2
⇒ sin-1 x = π/4
⇒ x = sin π/4 = 1/√2
What is the principal solutions of the equation
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 12 Detailed Solution
Download Solution PDFConcept:
The principal solutions of a trigonometric equation are those solutions that lie between 0 and 2π.
Formula:
General solution of tan(x) = tan(α) is given as;
x = nπ + α where α ∈ (-π/2 , π/2) and n ∈ Z.
Calculation:
Given,
⇒ tan(x) = tan(-π/6)
∴ α = -π/6
⇒ x = nπ + (-π/6) , n ∈ Z
Putting n = 1 and 2, we get -
x = 5π/6 and 11π/6
What is the value of cos (2tan-1 x + 2cot-1 x) ?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 13 Detailed Solution
Download Solution PDFConcept:
tan-1 x + cot-1 x =
Calculation:
To Find: Value of cos (2tan-1 x + 2cot-1 x)
cos (2tan-1 x + 2cot-1 x) = cos 2(tan-1 x + cot-1 x)
As we know, tan-1 x + cot-1 x =
cos (2tan-1 x + 2cot-1 x) = cos [2 ×
= cos π
= -1
In ΔABC, AB = 20 cm, BC = 21 cm and AC = 29 cm. What is the value of cot C + cosec C - 2tan A?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 14 Detailed Solution
Download Solution PDFGiven:
AB = 20 cm
BC = 21 cm
AC = 29 cm
Concept used:
Pythagoras' theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.
Calculation:
Using pythagoras theorem,
AC2 = AB2 + BC2
⇒ 292 = 202 + 212
ΔABC is a right angled triangle.
⇒ cot C = BC/AB = 21/20
⇒ cosec C = AC/AB = 29/20
⇒ tan A = BC/AB = 21/20
cot C + cosec C - 2tan A = 21/20 + 29/20 - 2 × 21/20
⇒ 8/20
⇒ 2/5
So, the value of cot C + cosec C - 2tan A = 2/5
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 15 Detailed Solution
Download Solution PDFConcept:
Calculation:
Given,
⇒
⇒
⇒
⇒
⇒
⇒
This is true for all x ∈ R