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

Calculation:

Given,

The function is \( y = 3 \sin^{-1} \left( \frac{x}{3} \right) \), and we are tasked with finding the derivative of y , i.e., \( \frac{dy}{dx} \).

The general derivative of \( \sin^{-1}(u) \) is:

\( \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \), where\( u = \frac{x}{3} \)

\( \frac{du}{dx} = \frac{1}{3} \)

Apply the chain rule to find the derivative of y :

\( \frac{dy}{dx} = 3 \times \frac{1}{\sqrt{1 - \left( \frac{x}{3} \right)^2}} \times \frac{1}{3} \)

\( \frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{x^2}{9}}} = \frac{3}{\sqrt{9 - x^2}} \)

Hence, the correct answer is Option 3. 

Calculation:

Given,

The function is \( y = \sin^{-1} \left( x - \frac{4x^3}{27} \right) \), and we are tasked with simplifying it.

\( y = \sin^{-1} \left( x - \frac{4x^3}{27} \right) \)

We recognize that we can factor the expression as:

\( x - \frac{4x^3}{27} = \left( \frac{x}{3} \right) \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right) \)

Thus, the function becomes:

\( y = \sin^{-1} \left( \frac{x}{3} \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right) \right) \)

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

\( \sin^{-1}(x) = 3 \sin^{-1} \left( \frac{x}{3} \right) \)

Thus, we can simplify the expression to:

\( y = 3 \sin^{-1} \left( \frac{x}{3} \right) \)

Hence, the correct answer is Option 4. 

Calculation:

Given,

\( \tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} \)

Use the addition formula for inverse tangents:

\( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \)

Using this for \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) \), we get:

\( \tan^{-1}\left(\frac{k + \frac{1}{2}}{1 - k \cdot \frac{1}{2}}\right) = \frac{\pi}{4} \)

Since \(\tan\left(\frac{\pi}{4}\right) = 1 \), we have:

\( \frac{k + \frac{1}{2}}{1 - \frac{k}{2}} = 1 \)

\( k + \frac{1}{2} = 1 - \frac{k}{2} \)

\( 2k + 1 = 2 - k \)

\( 2k + k = 2 - 1 \)

\( 3k = 1 \)

\( k = \frac{1}{3} \)

Hence, the correct answer is Option 3.

Calculation: 

Case I : x > 0

\(\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}\)

x = 2 - √3

Case II : x < 0 

\(\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}+\pi=\frac{\pi}{3}\)

\(x=\frac{-1}{\sqrt{3}} \Rightarrow \alpha=2\)

Hence, the correct answer is 2. 

Concept:

tan-1 x + cot-1 x = \(\rm \frac {\pi}{2}\)

Calculation:

As we know tan-1 x + cot-1 x = \(\rm \frac {\pi}{2}\)

∴ cot (tan-1 x + cot-1 x) = cot \(\rm \frac {\pi}{2}\)= 0

Concept:

\(\rm \cot \theta = \tan \left(\dfrac{\pi}{2}-\theta\right)\)​.

\(\tan^{-1} x = \dfrac{\pi}{2}-\cot^{-1} x\)

Calculation:

4 tan-1 x + cot‑1 x = π

\( 4\tan^{-1}x+\left(\dfrac{\pi}{2}-\tan^{-1}x\right)=\pi\)

\(3\tan^{-1}x=\pi-\dfrac{\pi}{2}=\dfrac{\pi}{2}\)

\( \tan^{-1}x=\dfrac{\pi}{6}\)

\(x=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}\).

Concept:

• \(\rm \tan^{-1}x + \tan^{-1}y=\tan^{-1}{x+y\over 1-xy}\)
• \(\rm cot^{-1}{x} = {\pi\over2}- \tan^{-1}{x}\)
• \(\rm 2tan^{-1}\ x =tan^{-1} ({\frac {2x}{1\ -\ x^2}})\)

Calculation:

S = \(\rm cot^{-1}{1\over3} - 2 \tan^{-1}{2\over3} \)

S = \(\rm \left[{\pi\over2}-\tan^{-1}{1\over3}\right] - \tan^{-1}{{2\times \frac{2}{3}}\over1-{(\frac{2}{3})^2}}\)

S = \(\rm \left[{\pi\over2}-\tan^{-1}{1\over3}\right] - \tan^{-1}{{\frac{4}{3}}\over1-{\frac{4}{9}}}\)

S = \(\rm {\pi\over2}-\tan^{-1}{1\over3} - \tan^{-1}{\frac{12}{5}}\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{12\over5}+\tan^{-1}{1\over3}\right]\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{{12\over5}+{1\over3}\over1-{12\over5}\times{1\over3}}\right]\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{41\over3}\right]\)

S = \(\rm \cot^{-1}{41\over3}\)

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

⇒ \(\rm -\dfrac{1}{4}\leq x \leq \dfrac{1}{4}\)

⇒ x ∈ \(\rm \left[-\dfrac{1}{4},\dfrac{1}{4}\right]\)

∴ The domain of the function is the closed interval \(\rm \left[-\dfrac{1}{4},\dfrac{1}{4}\right]\).

Concept:

sin^-1 x + cos^-1 x = \(\rm \frac{π}{2}\)  

Calculation:

sin^-1 x + sin^-1 y = \(\rm \frac{3π}{4}\) 

⇒ \(\rm \left ( \frac{π}{2} -cos^{-1}x\right ) + \left ( \frac{π}{2}- cos^{-1}y\right ) = \frac{3π}{4}\) 

⇒ π - ( cos^-1 x + cos^-1 y ) = \(\rm \frac{3π}{4}\)

⇒ cos^-1 x + cos^-1 y = \(\rm\frac{\pi}{4}\) 

The correct option is 2.

Concept:

\({\sin ^{ - 1}}\ (sin x) = x\) if \(x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\) but if \(x{ \notin }\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\), then use \(\sin x = \sin \left( {\pi - x} \right)\) to bring the value of x inside the principle branch.

Solution:

\({\sin ^{ - 1}}\left( {\sin \frac{{4\pi }}{5}} \right)\) but \(\frac{{4\pi }}{5}\notin{ }\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\)

So, use the relation,

\(\sin \frac{{4\pi }}{5} = \sin \left( {\pi - \frac{{4\pi }}{5}} \right)\)

\(= \sin \left( {\frac{\pi }{5}} \right)\)

So,

\({\sin ^{ - 1}}\left( {\sin \frac{{4\pi }}{5}} \right) = {\sin ^{ - 1}}\left( {\sin \frac{\pi }{5}} \right)\)

\( = \frac{\pi }{5}\)

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, \(\tan x=-\frac{1}{\sqrt{3}}\)

⇒ 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 = \(\rm \frac {π}{2}\)

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 = \(\rm \frac {π}{2}\)

cos (2tan-1 x + 2cot-1 x) = cos [2 × \(\rm \frac {π}{2}\) ]

= 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:

\(\rm tan^{-1}(x)+tan^{-1}({y})\) = \(\rm tan^{-1}\frac{x+y}{1-x\times y}\)

\(\rm cot^{-1}x=\) \(\rm tan^{-1}({1\over x})\)

Calculation:

Given, \(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{1-x\times \frac{1}{x}} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}({\infty}) = \frac{\pi}{2}\)

This is true for all x ∈ R