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

Concept:

tan-1 x + cot-1 x = 

Calculation:

As we know tan-1 x + cot-1 x = 

∴ cot (tan-1 x + cot-1 x) = cot = 0

Concept:

The maxima and minima of any function f(x) can be calculated by keeping it's first derivative as zero.

Now to find out whether it is a maxima or minima, we take it's double derivative now if the value is negative then it is a maxima and if it is positive it is a minima. For value equals to zero, it is unstable.

Solution: Given function,

To find out whether it is a maxima or minima, we take the double derivative.

Hence, 4>0. It is minima at x = 90°.

The correct answer is option 1.

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

sin (sin^-1 x) =x for -π/2 ≤ x ≤ π/2

We have, 

= sin^-1(sin( ))  ----- Since  ∈ [, ]

∴ sin-1(sin( )) = 

Additional InformationPrincipal Values of Inverse Trigonometric Functions:

Calculation:

Let, cot-1 = ϕ 

⇒ cot ϕ = 

⇒ sin ϕ =  = 

Let cos-1x = θ 

⇒ sec θ = 

⇒ tan θ = 

⇒ tan θ = 

⇒ tan θ = 

Now, (tan(cos-1x) = sin(cot-1)

⇒ (tan(tan-1) = sin(sin-1)

⇒  = 

⇒  = 2x

Squaring both sides, we get:

(1 - x²)5 = 4x2

⇒ 9x² = 5

⇒ x =  (∵ x > 0)

∴ The value of x is .

The correct answer is Option 2.

Explanation:

= 

=  (as cos(π + x) = -cos x)

=  (as cos(π - x) = -cos x)

= 

=  (cos^-1(cos x) = x if 0 ≤ x ≤ π)

Option (2) is true. 

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