## Mathematics MCQ on Inverse Trigonometry for JEE and Engineering Exam 2022

MCQ on Inverse Trigonometry: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject. In this post we are providing you MCQ on Inverse Trigonometry, which will be beneficial for you in upcoming JEE and Engineering Exams.

## MCQ on Inverse Trigonometry

Q1.  What is the value of sin-1(-x) for all x belongs to [-1, 1]?
a) sin-1(-x)/2
b) -sin-1(x)
c) sin-1(x)
d) 2sin-1(x)

b) -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)

Q2. 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)

b) π – cos-1(x)

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)

Q3. What is the value of sin-1(sin 6)?
a) -2π + 6
b) -2π – 6
c) 2π – 6
d) 2π + 6

a) -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

Q4. What will be the value of x + y + z if cos-1 x + cos-1 y + cos-1 z = 3π?
a) -2/3
b) 3
c) 1/3
d) -3

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.

Q5. Which value is similar to sin-1sin(6 π/7)?
a) coses-1(π/7)
b) sin-1(2π/7)
c) cos-1(π/7)
d) sin-1(π/7)

d) 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)

Q6. The given graph is for which equation? a) y = cotx
b) y = tan-1x
c) y = cosec-1x
d) y = cot-1x

d) y = cot-1x

Explanation: There are 2 curves. The black curve is the graph of y = cotx
The red curve is the graph for y = cot-1x
This curve does not pass through the origin but approaches to infinity in the direction of x axis only.
The part of the curve that lies in the (x, y) coordinate gradually meets to the x-axis.
This graph lies above +x axis and –x axis.

Q7. The given graph is for which equation? a)  y = sinx
b)  y = |sinx|
c)  y = |cosx|
d)  y = log|sinx|

b)  y = |sinx|

Explanation: The given form of equation can be written as, The green curve is the graph of y = sinx
The blue curve is the graph for y = |sinx|
As sinx is enclosed by a modulus so the curve that lies in the negative y axis will come to the positive y axis.

Q8. The given graph is for which equation? a) y = sin-1x
b) y = cosecx
c) y = secx
d) y = sinx

a) y = sin-1x

Explanation: The following graph represents 2 equations. The pink curve is the graph of y = sinx
The blue curve is the graph for y = sin-1x
This curve passes through the origin and approaches to infinity in both positive and negative axes.  