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Inverse Trigonometric Functions- Case Based Type Questions - Free


MCQ Practice Test & Solutions: Test: Inverse Trigonometric Functions- Case Based Type Questions (15 Questions)

You can prepare effectively for JEE Mathematics (Maths) Class 12 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Inverse Trigonometric Functions- Case Based Type Questions". These 15 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 15

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Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 1

Direction: Read the following text and answer the following questions on the basis of the same:

Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.

∠CAB = α =

Detailed Solution: Question 1

In Δ BDA

sin α = BD/AB

AB2 = AD2 + BD2

= (30√3)2 + 302

= 602

AB = 60 m

Now, sinα = 30/60

Sinα = ½

I.e. ∠CAB = α = sin-1(½)

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 2

Direction: Read the following text and answer the following questions on the basis of the same:

Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.

Domain and Range of cos−1 x =

Detailed Solution: Question 2

Since cos x is defined at x = 0, and x = π

Domain of cos−1 x includes -1 and 1

Range of cos−1 x also includes 0 and π

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 3

Direction: Read the following text and answer the following questions on the basis of the same:

In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan-1(½) and tan-1(⅓) respectively.

The value of sin A is _______.

Detailed Solution: Question 3

A = tan-1(½)

⇒ tan A = ½

∴ sin A = 1/√5

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 4

Direction: Read the following text and answer the following questions on the basis of the same:

In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan-1(½) and tan-1(⅓) respectively.

The third angle, ∠C = _______.

Detailed Solution: Question 4

∠C = π – (A + B)

= π - (π/4)

= 3π/4

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 5

Direction: Read the following text and answer the following questions on the basis of the same:

The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.

Principal value of sin–1 (½) is

Detailed Solution: Question 5

sin-1(½) = y

sin y = ½

Principal value branch of sin–1 is (-π/2, π/2)

and sin(π/6) = ½

⇒ Principal value of sin-1(½) is π/6.

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 6

Direction: Read the following text and answer the following questions on the basis of the same:

The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.

Principal value of 2cos–1 (1) + 5tan–1 (1) is:

Detailed Solution: Question 6

2cos-1(1) + 5tan-1(1)

= 2 x 0 + 5 x π/4

= 5π/4

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 7

Direction: Read the following text and answer the following questions on the basis of the same:

Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.

∠CAB = α =

Detailed Solution: Question 7

In Δ BDA

Cosα = AD/AB

Cosα = 30√3/60

α = cos-1(√3/2)

∴∠CAB = α = cos-1(√3/2)

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 8

Direction: Read the following text and answer the following questions on the basis of the same:

In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan-1(½) and tan-1(⅓) respectively.

cos(A + B + C) = _______.

Detailed Solution: Question 8

Since ABC is a triangle,

∴ A + B + C = 180°

cos (A + B + C) = cos 180°

= -1

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 9

Direction: Read the following text and answer the following questions on the basis of the same:

The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.

Principal value of tan–1 (1)

Detailed Solution: Question 9

tan-1(1) =tan-1(tan(π/4))

= π/4

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 10

Direction: Read the following text and answer the following questions on the basis of the same:

Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.

∠BCA = β =

Detailed Solution: Question 10

DC = AC - AD

= 40√3 - 30√3

= 10√3 m

In ΔBDC

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 11

Direction: Read the following text and answer the following questions on the basis of the same:

In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan-1(½) and tan-1(⅓) respectively.

If B = cos–1 x, then x = _______.

Detailed Solution: Question 11

Given B = tan-1(⅓)

⇒ tan B = ⅓

∴ cos B = 3/√10

B = cos-1(3/√10)

⇒ x = 3/√10

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 12

Direction: Read the following text and answer the following questions on the basis of the same:

The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.

Principal value of cot-1(√3) is :

Detailed Solution: Question 12

cot -1(√3) =cot-1 (cot (π/6) = π/6

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 13

Direction: Read the following text and answer the following questions on the basis of the same:

Two men on either side of a temple of 30 metres high observe its top at the angles of elevation α and β respectively. (as shown in the figure above). The distance between the two men is 40√3 metres and the distance between the first person A and the temple is 30√3 meters.

∠ABC =

Detailed Solution: Question 13

Since,

Sin α = ½

i.e., sin α = sin 300 [∵ sin 300 = ½]

∴ α = 300

We have tan β = √3

tan β = tan 600

∴ β = 600

Now In ΔABC

∠ABC + ∠BCA + ∠CAB = 1800

∠ABC + 600 + 300 = 1800

∠ABC = 900

∴ ∠ABC = π/2

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 14

Direction: Read the following text and answer the following questions on the basis of the same:

In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan-1(½) and tan-1(⅓) respectively.

If A = sin–1x; then the value of x is:

Detailed Solution: Question 14

A = tan-1 (½)

⇒ tan A = ½

∴ Sin A = 1/√5

A = sin-1(1/√5)

⇒ x = 1/√5

Test: Inverse Trigonometric Functions- Case Based Type Questions - Question 15

Direction: Read the following text and answer the following questions on the basis of the same:

The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.

Principal value of sin -1(1) + sin-1(1/√2) is

Detailed Solution: Question 15

sin-1(1) + sin-1(1/√2) = π//2 + π/4

= 3π/4

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