Inverse Trigonometric Functions MCQ Quiz - Objective Question with Answer for Inverse Trigonometric Functions - Download Free PDF
Last updated on Apr 22, 2025
Latest Inverse Trigonometric Functions MCQ Objective Questions
Inverse Trigonometric Functions Question 1:
Find value of cot (tan-1 x + cot-1 x)
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 1 Detailed Solution
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
Inverse Trigonometric Functions Question 2:
\(f(x)=1+2\sin x+3\cos ^2x, \left(0\le x\le \frac{2x}{3}\right)\) is
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 2 Detailed Solution
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,
\(f(x)=1+2\sin x+3\cos ^2x, \left(0\le x\le \frac{2x}{3}\right) \)
\(f^{\prime}(x)=2\cos x-6\sin x cos x\\ f^{\prime}(x)=0 \\ 2\cos x-6\sin x cos x=0 \\ 2\cos x(1-3\sin x)=0 \\ \implies \text{either} \cos x=0 \ \text{or} \ (1-3\sin x)=0 \\ x=\sin ^{-1}(\frac{1}{3}),\frac{\pi}{2}\)
To find out whether it is a maxima or minima, we take the double derivative.
\(f^{\prime \prime}(x)=-2\sin x-6 \cos 2x\\ \text{for value} \frac{\pi}{2},\\ f^{\prime \prime}(x)=-2\sin ( \frac{\pi}{2})-6 \cos (2 \cdot \frac{\pi}{2})\\ f^{\prime \prime}(x)=-2\sin ( \frac{\pi}{2})-6 \cos \pi\\ f^{\prime \prime}(x)=-2+6 =4\)
Hence, 4>0. It is minima at x = 90°.
The correct answer is option 1.
Inverse Trigonometric Functions Question 3:
The principal value of \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\) is
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 3 Detailed Solution
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, \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\)
= sin-1(sin( \(\frac{-\pi}{3}\))) ----- Since \(\frac{-\pi}{3}\) ∈ [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)]
∴ sin-1(sin( \(\frac{-\sqrt{3}}{2}\))) = \(\frac{-\pi}{3}\)
Additional InformationPrincipal Values of Inverse Trigonometric Functions:
Function |
Domain |
Range of Principal Value |
sin-1 x |
[-1, 1] |
[-π/2, π/2] |
cos-1 x |
[-1, 1] |
[0, π] |
csc-1 x |
R - (-1, 1) |
[-π/2, π/2] - {0} |
sec-1 x |
R - (-1, 1) |
[0, π] - {π/2} |
tan-1 x |
R |
(-π/2, π/2) |
cot-1 x |
R |
(0, π) |
Inverse Trigonometric Functions Question 4:
If (tan(cos-1x) = sin(cot-1\(\frac{1}{2}\)), then x is equal to
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 4 Detailed Solution
Calculation:
Let, cot-1\(\frac{1}{2}\) = ϕ
⇒ cot ϕ = \(\frac{1}{2}\)
⇒ sin ϕ = \(\frac{1}{\sqrt{1+\cot^2\phi}}\) = \(\frac{2}{\sqrt{5}}\)
Let cos-1x = θ
⇒ sec θ = \(\frac{1}{x}\)
⇒ tan θ = \(\sqrt{\sec^2\theta-1}\)
⇒ tan θ = \(\sqrt{\frac{1}{x^2}-1}\)
⇒ tan θ = \(\frac{\sqrt{1-x^2}}{x}\)
Now, (tan(cos-1x) = sin(cot-1\(\frac{1}{2}\))
⇒ (tan(tan-1\(\frac{\sqrt{1-x^2}}{x}\)) = sin(sin-1\(\frac{2}{\sqrt{5}}\))
⇒ \(\frac{\sqrt{1-x^2}}{x}\) = \(\frac{2}{\sqrt{5}}\)
⇒ \(\sqrt{(1-x^2)5}\) = 2x
Squaring both sides, we get:
(1 - x2)5 = 4x2
⇒ 9x2 = 5
⇒ x = \(\frac{\sqrt{5}}{3}\) (∵ x > 0)
∴ The value of x is \(\frac{\sqrt{5}}{3}\).
The correct answer is Option 2.
Inverse Trigonometric Functions Question 5:
Principal value of \(\rm \cos^{-1}(\cos \left(\frac{5\pi}{4}\right))\) is
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 5 Detailed Solution
Explanation:
\(\rm \cos^{-1}(\cos \left(\frac{5π}{4}\right))\)
= \(\rm \cos^{-1}(\cos \left(π+\frac{π}{4}\right))\)
= \(\rm \cos^{-1}(-\cos \left(\frac{π}{4}\right))\) (as cos(π + x) = -cos x)
= \(\rm \cos^{-1}(\cos \left(π-\frac{π}{4}\right))\) (as cos(π - x) = -cos x)
= \(\rm \cos^{-1}(\cos \left(\frac{3π}{4}\right))\)
= \(\frac{3π}{4}\) (cos-1(cos x) = x if 0 ≤ x ≤ π)
Option (2) is true.
Top Inverse Trigonometric Functions MCQ Objective Questions
If 4 tan-1 x + cot‑1 x = π, then x equals:
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 6 Detailed Solution
Download Solution PDFConcept:
\(\rm \cot \theta = \tan \left(\dfrac{\pi}{2}-\theta\right)\).
\(\tan^{-1} x = \dfrac{\pi}{2}-\cot^{-1} x\)
Calculation:
4 tan-1 x + cot‑1 x = π
\( 4\tan^{-1}x+\left(\dfrac{\pi}{2}-\tan^{-1}x\right)=\pi\)
\(3\tan^{-1}x=\pi-\dfrac{\pi}{2}=\dfrac{\pi}{2}\)
\( \tan^{-1}x=\dfrac{\pi}{6}\)
\(x=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}\).
\(\rm cot^{-1}{1\over3} - 2 \tan^{-1}{2\over3} = ?\)
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 7 Detailed Solution
Download Solution PDFConcept:
- \(\rm \tan^{-1}x + \tan^{-1}y=\tan^{-1}{x+y\over 1-xy}\)
- \(\rm cot^{-1}{x} = {\pi\over2}- \tan^{-1}{x}\)
- \(\rm 2tan^{-1}\ x =tan^{-1} ({\frac {2x}{1\ -\ x^2}})\)
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 = \(\rm \cot^{-1}{41\over3}\)
The domain of sin-1 4x is:
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 8 Detailed Solution
Download Solution PDFConcept:
- 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 \(\rm \left[-\dfrac{1}{4},\dfrac{1}{4}\right]\).
If sin-1 x + sin-1 y = \(\rm \frac{3\pi}{4}\) , then cos-1 x + cos-1 y = ? .
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 9 Detailed Solution
Download Solution PDFConcept:
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.
Find the value of \({\sin ^{ - 1}}\left( {\sin \frac{{4\pi }}{5}} \right)\)
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 10 Detailed Solution
Download Solution PDFConcept:
\({\sin ^{ - 1}}\ (sin x) = x\) 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,
\({\sin ^{ - 1}}\left( {\sin \frac{{4\pi }}{5}} \right) = {\sin ^{ - 1}}\left( {\sin \frac{\pi }{5}} \right)\)
\( = \frac{\pi }{5}\)
If 3 sin-1 x + cos-1 x = π, then find the value of x?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 11 Detailed Solution
Download Solution PDFConcept:
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
What is the principal solutions of the equation \(\tan x=-\frac{1}{\sqrt{3}}\)?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 12 Detailed Solution
Download Solution PDFConcept:
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
What is the value of cos (2tan-1 x + 2cot-1 x) ?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 13 Detailed Solution
Download Solution PDFConcept:
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
In ΔABC, AB = 20 cm, BC = 21 cm and AC = 29 cm. What is the value of cot C + cosec C - 2tan A?
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 14 Detailed Solution
Download Solution PDFGiven:
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
\(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\) holds, when
Answer (Detailed Solution Below)
Inverse Trigonometric Functions Question 15 Detailed Solution
Download Solution PDFConcept:
\(\rm tan^{-1}(x)+tan^{-1}({y})\) = \(\rm tan^{-1}\frac{x+y}{1-x\times y}\)
\(\rm cot^{-1}x=\) \(\rm tan^{-1}({1\over x})\)
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