Inverse Trigonometric Functions MCQ Quiz - Objective Question with Answer for Inverse Trigonometric Functions - Download Free PDF

Calculation:

Given,

The function is , and we are tasked with finding the derivative of y , i.e., .

The general derivative of is:

, where

Apply the chain rule to find the derivative of y :

Hence, the correct answer is Option 3. 

Calculation:

Given,

The function is , and we are tasked with simplifying it.

We recognize that we can factor the expression as:

Thus, the function becomes:

Recognizing that this fits a known identity for inverse trigonometric functions:

Thus, we can simplify the expression to:

Hence, the correct answer is Option 4. 

Calculation:

Given,

Use the addition formula for inverse tangents:

Using this for , we get:

Since , we have:

Hence, the correct answer is Option 3.

Concept:

• The question uses inverse trigonometric identities of tan^-1 and sin^-1.
• The identity for sin(2φ) is applied to simplify the equation:
  • sin(2φ) = 2tanφ / (1 + tan²φ)
• The domain restrictions are carefully analyzed for different cases of φ in (-π, π).

Calculation:

Let = tan φ

⇒ 2tan θ / (1 + tan²θ) = sin 2φ

⇒ θ = tan^-1(2 tan θ) − ½ sin^-1(sin 2φ)

Let 2φ ∈

⇒ φ ∈

⇒ tan φ ∈ (−1, 1), tan θ ∈ (−3, 3)

Case 1:

Let 2φ =

⇒ tan θ = tan^-1(2 tan θ) − φ

⇒ tan 2τ = tan θ / 3

⇒ τ = tan^-1((1/3), 0, 1, −1)

⇒ tan θ = 0, 1, −1, 3/2

⇒ θ ∈ 0, 1, −1 and all are in domain

Case 2:

Let 2φ ∈

⇒ θ = tan^-1(2 tan θ) − (π − 2φ)

⇒ tan 2τ = cot θ = 1 / tan θ

⇒ θ = tan^-1(3/2)

⇒ τ = tan^-1(3/2)

So, θ = tan^-1(3/2), −2π/3, −π/3 ∴ θ = 1 = φ

Case 3:

Let 2φ ∈

⇒ θ = tan^-1(2 tan θ) − (−π − 2φ)

⇒ Proceed similarly to other cases.

∴ Total valid θ values are 3.

Calculation: 

Case I : x > 0

x = 2 - √3

Case II : x

Hence, the correct answer is 2. 

Concept:

​.

Calculation:

4 tan-1 x + cot‑1 x = π

.

Concept:

Calculation:

S = 

S = 

S = 

S = 

S = 

S = 

S = 

S = 

Concept:

• 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 .

Concept:

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.

Concept:

 if  but if , then use  to bring the value of x inside the principle branch.

Solution:

 but 

So, use the relation,

So,

Concept:

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

Concept:

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

Concept:

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

Given:

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

Concept:

 = 

Calculation:

Given, 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

This is true for all x ∈ R