Inverse Trigonometric Functions Class 12 Mathematics Exam Questions

Please refer to Inverse Trigonometric Functions Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 12 Mathematics Exam Questions Inverse Trigonometric Functions

Class 12 Mathematics students should read and understand the important questions and answers provided below for Inverse Trigonometric Functions which will help them to understand all important and difficult topics.

Question. Determine the principal value of cos^-1( -1/2).
Answer. Let us assume that, y = cos^-1( -1/2)
We can write this as:
cos y =- 1/2
cos y = cos (2π/3).
Thus, the Range of the principal value of cos^-1 is [0, π ].
Therefore, the principal value of cos^-1( -1/2) is 2π /3.

Question. Find the value of cot (tan^-1 α + cot^-1 α).
Answer. Given that: cot (tan^-1 α + cot^-1 α)
= cot (π/2) (since, tan^-1 x + cot^-1 x = 𝜋/2)
= cot (180°/2) ( we know that cot 90° = 0 )
= cot (90°)
= 0
Therefore, the value of cot (tan-1 α + cot^-1 α) is 0.

Question. The value of tan^-1 √3 – sec^-1(–2) is equal to:
(A) π (B) – π/3 (C) π/3 (D) 2π/3
Answer. Now, solve the first part of the expression: tan^-1 √3
Let us take y = tan^-1√3
This can be written as:
tan y = √3
Now, use the trigonometry table to find the radian value
tan y = tan (π/3)
Thus, the range of principal value of tan^-1 is (−π/2, π/2)
Therefore, the principal value of tan^-1√3 is π/3.
Now, solve the second part of the expression: sec^-1(–2)
Now, assume that y = sec^-1 (–2)
sec y = -2
sec y = sec (2π/3)
We know that the principal value range of sec^-1 is [0,π] – {π/2}
Therefore, the principal value of sec^-1 (–2) = 2π/3
Now we have:
tan^-1(√3) = π/3
sec^-1 (–2) = 2π/3
Now, substitute the values in the given expression:
= tan^-1 √3 – sec^-1 (−2)
= π/3 − (2π/3)
= π/3 − 2π/3
= (π − 2π)/3
= – π/3
Hence, the correct answer is an option (B)

Question. Prove that sin^-1 (3/5) – sin^-1 (8/17) = cos^-1 (84/85).
Answer. Let sin^-1 (3/5) = a and sin^-1 (8/17) = b
Thus, we can write sin a = 3/5 and sin b = 8/17
Now, find the value of cos a and cos b
To find cos a:
Cos a = √[1 – sin² a]
= √[1 – (3/5)² ]
= √[1 – (9/25)]
= √[(25-9)/25]
= 4/5
Thus, the value of cos a
= 4/5
To find cos b:
Cos b = √[1 – sin² b]
= √[1 – (8/17)2 ]
= √[1 – (64/289)] = √[(289-64)/289] = 15/17
Thus, the value of cos b = 15/17
We know that cos (a- b) = cos a cos b + sin a sin b
Now, substitute the values for cos a, cos b, sin a and sin b in the formula, we get:
cos (a – b) = (4/5)x (15/17) + (3/5)x(8/17)
cos (a – b) = (60 + 24)/(17x 5)
cos (a – b) = 84/85
(a – b) = cos^-1 (84/85)
Substituting the values of a and b sin^-1 (3/5)- sin^-1 (8/7) = cos^-1 (84/85)
Hence proved.

Question. Find the value of cos^-1 (1/2) + 2 sin^-1 (1/2).
Answer. First, solve for cos^-1 (1/2):
Let us take, y = cos^-1 (1/2)
This can be written as:
cos y = (1/2)
cos y = cos (π /3).
Thus, the range of principal value of cos^-1 is [0, π ]
Therefore, the principal value of cos^-1 (1/2) is π/3.
Now, solve for sin^-1 (1/2):
Let y = sin^-1 (1/2)
sin y = 1/2
sin y = sin ( π/6)
Thus, the range of principal value of sin^-11 is [(-π)/2, π/2 ]
Hence, the principal value of sin^-1 (1/2) is π/6.
Now we have cos^-1 (1/2) = π/3 & sin^-1 (1/2) = π/6
Now, substitute the obtained values in the given formula, we get:
= cos^-1 (1/2) + 2sin^-1 (1/2)
= π /3 + 2( π/6)
= π/3 + π/3
= ( π+π )/3
= 2π /3
Thus, the value of cos^-1 (1/2) + 2 sin^-1 (1/2) is 2π /3.

Question. Find the principal value of cos ^-1( √3 / 2 ) .
Answer. Let cos^-1 (√3 / 2 ) = y . Then, cos y = √3 / 2.
The range of the principal values of y = cos^-1 x is [0, π ].
So, let us find y in [0, π ] such that cos y = √3 / 2
But, cos π/6 = √3/2 and π/6 ∈ [0,π ]. Therefore, y = π /6
Thus, the principal value of cos^-1 (√3/2 ) is π/6 .

Question. Find the principal value of tan^-1 (√3)
Answer. Let tan^-1 (√3) = y .
Then, tan y = √3.
Thus, y = π/3 . Since π/3 ∈ ( – π/2 , π/2 ) .
Thus, the principal value of tan-1(√3) is π/3.

Question. Find (i) tan^-1 (-√3 )
(ii) tan^-1( tan (3π / 5))
(iii) tan (tan^-1 (2019))
Answer.

(iii) Since tan (tan^-1 x) = x, x ∈ R , we have tan (tan^-1 (2019)) = 2019.

Question. Find the value of tan^-1 (-1) + cos^-1 ( 1/2 ) + sin^-1 ( – 1/2 ) .
Answer.  Let tan^-1 (-1) = y . Then, tan y = -1 = – tan π/4 = tan ( – π/4 )

Question. Prove that tan (sin^-1 x) =

Answer. If x = 0 , then both sides are equal to 0. ………..(1)
Assume that 0 < x < 1.
Let θ = sin-1 x . Then 0< θ < π/2 . Now, sin θ = x/1 gives tan θ =

Hence, tan (sin^-1x) =

Assume that -1 < x < 0. Then, θ = sin^-1x gives – π/2 < θ < 0. Now, sinθ = x/1 gives tanθ =

In this case also, tan (sin-1 x) =

Equations (1), (2) and (3) establish that tan (sin^-1 x) =

Question. Find

Answer. It is known that cos-1 x : [-1, 1] → [0, π ] is given by
Cos-1 x = y if and only if x = cos y for -1 ≤ x ≤ 1 and 0 ≤ y ≤ π .
Thus, we have

Question. Find the principal value of
(i) cosec^-1 (-1) (ii) sec^-1 (-2) .
Answer. (i) Let cosec-1 (-1) = y . Then, cosec y = -1
Since the range of principal value branch of y= cosec-1 x is [- π/2 , π/2] \ {0} and

Thus, the principal value of cosec-¹ (-1) is – π/2 .
(ii) Let y = sec^-1 (-2) . Then, sec y = -2 .
By definition, the range of the principal value branch of y = sec^-1 x is [0,π ]\ {π /2} .
Let us find y in [0,π ] – {π/2} such that sec y = -2 .
But, sec y = −2 ⇒ cos y = − 1/2 .
Now, cos y =- 1/2 = -cos π/3 = cos (π – π/3 ) = cos 2π/3 . Therefore, y = 2π/3 .
Since 2π/3 ∈ [0, π ] \ {π/2 } , the principal value of sec^-1 (-2) is 2π/3 .

Question. Find the value of sec^-1(- 2√3 / 2)
Answer.

Question. If cot^-1 ( 1/7 ) = θ , find the value of cos θ .
Answer. By definition, cot^-1 x ∈ (0, π) .
Therefore, cot^-1 (1/7) = θ implies cot θ ∈ (0,π ) .

But cot^-1 ( 1/7 ) = θ implies cot θ = 1/7 and hence tan θ = 7 and θ is acute.
Using tan θ = 7/1 , we construct a right triangle as shown . Then, we have, cosθ = 1/ 5√2 .

Question.

Answer.

We construct a right triangle with the given data.
From the triangle, secα = x/1 = x . Thus, α = sec^-1 x .

Question.

Answer.

Question.

Answer.

Question.

Answer

Question. Solve tan^-1 2x + tan^-1 3x = π/4 , if 6x² < 1.
Answer. Now, tan^-1 2x + tan^-1 3x = tan^-1 (2x + 3x/1 – 6x) since 6x² <1 .
So, tan^-1(5x/1 – 6x) = π/4 , which implies 5x/ [1- 6x²] = tan π /4 = 1
Thus, 1- 6x² = 5x , which gives 6x² + 5x -1 = 0
Hence, x = 1/6 , -1 . But x = -1 does not satisfy 6x² <1 .
Observe that x = -1 makes the left side of the equation negative whereas the right side is a positive number. Thus, x = -1 is not a solution.

Question.

Answer.

CASE STUDY:

The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the
ground level. For the viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:

Question. Measure of ∠𝐶𝐴𝐵 =
a. tan⁻¹⁽2)
b. tan⁻¹( 1/2)
c. tan⁻¹( 1)
d. tan⁻¹( 3)

Question. 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓∠𝐷𝐴𝐵 =
a. tan⁻¹(3/4)
b. tan⁻¹(3)
c. tan⁻¹(4/3)
d. tan⁻¹(4)

Question. 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 ∠𝐸𝐴𝐵=
a. tan⁻¹(11)
b. tan⁻¹(3)
c. tan⁻¹(2/11)
d. tan⁻¹(11/2)

Question. 𝐴| Is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠𝐶𝐴𝐵 and ∠𝐶𝐴′𝐵
Is
a. tan⁻¹(1/2)
b. tan⁻¹ (1/8)
c. tan⁻¹( 2/5)
d. tan⁻¹(11/21)

Question. Domain and Range of ^tan−1𝑥=
a. 𝑅⁺, (−𝜋/2,𝜋/2)
b. 𝑅⁻, (−𝜋/2,𝜋/2)
c. R , (−𝜋/2,𝜋/2)
d. R , (0 ,𝜋/2)