Case Based MCOs
1. Two men on either side of a temple of 30 m high observe its top at the angles of elevation 𝛼 and
𝛽 respectively. (as shown in the figure below).
The distance between two men is 40√3 m and the distance between the first person 𝐴 and the temple
is 30√3 m. Based on the above information answer the following questions.
[CBSE Question Bank]
(i) ∠𝐶𝐴𝐵 = 𝛼 is
2
(a) sin−1 ( 3)
√
1
(b) sin−1 (2)
(c) sin−1 (2)
√3
(d) sin−1 ( 2 )
(ii) ∠𝐶𝐴𝐵 = 𝛼 is
1
(a) cos−1 (5)
2
(b) cos−1 ( )
5
−1 √3
(c) cos ( )
2
4
(d) cos−1 (5)
(iii) ∠𝐵𝐶𝐴 = 𝛽 is
1
(a) tan−1 (2)
(b) tan−1 (2)
1
(c) tan−1 ( )
√3
(d) tan−1 (√3)
(iv) ∠𝐴𝐵𝐶 is
𝜋
(a) 4
𝜋
(b) 6
𝜋
(c) 2
𝜋
(d) 3
1
(v) Domain and Range of cos−1 𝑥 is
(a) (−1,1), (0, 𝜋)
(b) [−1,1], (0, 𝜋)
(c) [−1,1], [0, 𝜋]
𝜋 𝜋
(d) (−1,1), [− 2 , 2 ]
Solution
Given figure,
𝐵𝐷 30 1
(i) In △ 𝐴𝐵𝐷, tan 𝛼 = = =
𝐴𝐷 30√3 √3
⇒ 𝛼 = 30∘
1 1
∴ sin 𝛼 = sin 30∘ = ⇒ 𝛼 = sin−1 ( )
2 2
Ans-b
(ii) In △ 𝐴𝐵𝐷,
√3 √3
cos 𝛼 = cos 30∘ = ⇒ 𝛼 = cos −1 ( )
2 2
Ans-c
𝐵𝐷 30
(iii) In △ 𝐵𝐷𝐶, tan 𝛽 = 𝐶𝐷 = 10 = √3
√3
⇒ 𝛽 = tan−1 (√3)
Ans-d
1 𝜋
(iv) Since, 𝛼 = sin−1 (2) = 6 ,
𝜋
𝛽 = tan−1 (√3) =
3
𝜋 𝜋
∴ ∠𝐴𝐵𝐶 = 𝜋 − ( + )
6 3
𝜋 𝜋
= 𝜋 − =
2 2
Ans-c
2
(v) We know that, domain and range of cos−1 𝑥 are [−1,1] and [0, 𝜋] respectively.
Ans-b
2. As the trigonometric functions are periodic functions, so these functions are many-one.
Trigonometric functions are not one-one and onto over their natural domain and range, so their
inverse do not exist. But if we restrict their domain and range, then their inverse may exists.
Based on the above information, answer the following questions.
𝜋 3𝜋
(i) The value of tan−1 (−1) in the interval ( 2 , 2
)
𝜋
(a)
4
3𝜋
(b) 4
5𝜋
(c) 4
7𝜋
(d) 4
𝜋 1
(ii) sin ( 3 − sin−1 (2)) is equal to
1
(a)
2
1
(c)
4
1
(b) 3
(d) 1
(iii) The value of tan−1 (√3) + cot −1 (√3) +tan−1 (cos(0)) is
𝜋
(a)
6
𝜋
(b)
12
3𝜋
(c)
4
𝜋
(d) 3
√3
(iv) The value of tan−1 [2sin (2cos−1 ( 2 ))] is
𝜋
(a) 3
2𝜋
(b) 3
𝜋
(c) − 3
𝜋
(d)
6
√3
(v) The principal value of cos−1 ( ) is
2
𝜋
(a) 6
𝜋
(b) 4
𝜋
(c) 3
𝜋
(d) 2
3
Solution
(i) Let 𝑦 = tan−1 (−1)
⇒ tan 𝑦 = −1
3𝜋 3𝜋 𝜋 3𝜋
⇒ tan 𝑦 = tan [∵ ∈ ( , )]
4 4 2 2
3𝜋
⇒ 𝑦 =
4
Ans-b
𝜋 1
(ii) sin [ 3 − sin−1 (2)]
𝜋 𝜋 1 𝜋
= sin [ − ] [∵ sin−1 ( ) = ]
3 6 2 6
𝜋 1
= sin =
6 2
Ans-a
(iii) tan−1 (√3) + cot −1 (√3) + tan−1 (cos(0))
𝜋 𝜋
= + + tan−1 (1)
3 6
𝜋 𝜋 2𝜋 + 𝜋 3𝜋
= + = =
2 4 4 4
Ans-c
√3
(iv) Given, tan−1 [2sin (2cos −1 ( ))]
2
𝜋 𝜋
= tan−1 [2sin (2 × )] = tan−1 (2sin )
6 3
√3 𝜋
= tan−1 (2 × ) = tan−1 √3 =
2 3
Ans-a
√3 √3
(v) Let cos−1 ( 2 ) = 𝜃 ⇒ cos 𝜃 = 2
We know that, principal value branch of cos−1 is [0, 𝜋].
√3 𝜋 𝜋
∴ cos 𝜃 = = cos ⇒ 𝜃 =
2 6 6
√3 𝜋
Hence, cos−1 ( 2 ) = 6
∈ [0, 𝜋]
Ans-a
4
3. Ritika, a student of class XII is teaching a new topic on existence of a function in their domain.
For six inverse trigonometric functions, she is teaching about their domain and range.
Based on domain and range of the inverse trigonometric functions, answer the following questions.
1
(i) The value of sin−1 𝑥 at 𝑥 = 2 is
(a) does not exist
(b) a finite value
𝜋
(c) 2
𝜋
(d)
4
(ii) The value of cos−1 𝑥 at 𝑥 = 2 is
(a) does not exist
(b) a finite value
𝜋
(c) 3
𝜋
(d)
2
(iii) The value of tan−1 𝑥 at 𝑥 = 6 is
(a) does not exist
(b) a finite value
𝜋
(c) 4
𝜋
(d) 3
1
(iv) The value of sec −1 (3) is
(a) does not exist
(b) a finite value
𝜋
(c) 4
𝜋
(d) 3
(v) The value of cosec −1 (√2) is
(a) does not exist
𝜋
(b) 3
𝜋
(c) 4
𝜋
(d) 6
5
Solution
(i) We know that,
The domain of sin−1 𝑥 is [−1,1]
1 1 1 𝜋
At 𝑥 = ∈ [−1,1]; sin−1 ( ) exists and it is a finite value which is sin−1 ( ) = .
2 2 2 6
Ans-b
(ii) We know that,
The domain of cos−1 𝑥 is [−1,1]
𝑥 = 2 ∉ [−1,1]
Hence, cos−1 (2) does not exist.
Ans-a
(iii) We know that,
The domain of tan−1 𝑥 is (−∞, ∞).
𝜋
It means tan−1 𝑥 exists for all real numbers. At 𝑥 = 6, value of tan−1 (6) is a finite value but not 4
𝜋
or .
3
Ans-b
(iv) We know that,
1
The domain of sec −1 𝑥 is (−∞, −1] ∪ [1, ∞) ∵ 3 ∉ domain of sec −1 𝑥.
1
∴ sec −1 ( ) does not exist.
3
Ans-a
(v) We know that,
The domain of cosec −1 𝑥 is (−∞, −1] ∪ [1, ∞).
∵ √2 ∈ domain of cosec −1 𝑥.
∴ cosec −1 𝑥 exists at 𝑥 = √2
𝜋
and cosec −1 (√2) = 4
Ans-c
