Class 12 Maths Chapter 2 Inverse Trigonometric Functions MCQ

This set of Class 12 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Inverse Trigonometric Functions”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. \(tan^{-1}\sqrt{3}+sec^{-1}⁡2 – cos^{-1}⁡1\) is equal to ________
a) 0
b) \(\frac{2π}{3}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{4}\)
View Answer

Answer: b
Explanation: \(tan^{-1}\sqrt{3}=\frac{π}{3}, sec^{-1}⁡2=\frac{π}{3}, cos^{-1}⁡1=0\)
∴\(tan^{-1}\sqrt{3}+sec^{-1}⁡2 -cos^{-1}⁡1=\frac{π}{3}+\frac{π}{3}-0\)
=\(\frac{2π}{3}\).

2. What is the principle value of \(sec^{-1}⁡(\frac{2}{\sqrt{3}})\).
a) \(\frac{π}{6}\)
b) \(\frac{π}{3}\)
c) \(\frac{π}{4}\)
d) \(\frac{π}{2}\)
View Answer

Answer: a
Explanation: Let \(sec^{-1}⁡(\frac{2}{\sqrt{3}})\)=y
sec⁡ y=\(\frac{2}{\sqrt{3}}\)
sec⁡ y=sec \(⁡\frac{π}{6}\)
⇒y=\(\frac{π}{6}\)

3. What is the value of \(tan^1⁡\frac{1}{√3}-sin^{-1}⁡1+ cos^{-1}\frac{⁡1}{2}\) is ________
a) 2π
b) \(⁡\frac{π}{2}\)
c) π
d) 0
View Answer

Answer: c
Explanation: \(tan^{-1}⁡\frac{1}{\sqrt{3}}=\frac{π}{6},sin^{-1}⁡1=\frac{π}{2}, cos^{-1}\frac{⁡1}{2}=\frac{π}{3}\)
\(tan^1⁡\frac{1}{\sqrt{3}}-sin^{-1}⁡1+ cos^{-1}⁡\frac{1}{2}=\frac{π}{6}+\frac{π}{2}+\frac{π}{3}=\frac{π+3π+2π}{6}=\frac{6π}{6}=π\)

4. [-1, 1] is the domain for which of the following inverse trigonometric functions?
a) sin^-1⁡x
b) cot^-1⁡x
c) tan^-1⁡x
d) sec^-1⁡x
View Answer

Answer: a
Explanation: [-1, 1] is the domain for sin^-1⁡x.
The domain for cot^-1⁡x is (-∞,∞).
The domain for tan^-1⁡x is (-∞,∞).
The domain for sec^-1⁡x is (-∞,-1]∪[1,∞).

5. The domain of sin^-1⁡(3x) is equal to _______
a) [-1, 1]
b) \([\frac{-1}{3}, \frac{1}{3}]\)
c) [-3, 3]
d) [-3π, 3π]
View Answer

Answer: b
Explanation: The domain of y=sin^-1⁡x is -1≤x≤1.
∴the domain of y=sin^-1⁡3x is-1≤3x≤1
⇒ \([\frac{-1}{3} ≤ x ≤ \frac{1}{3}]\)
Hence, \([\frac{-1}{3}, \frac{1}{3}]\).

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6. What is the value of 5 \(cos^{-1}\frac{⁡1}{2} + 7 sin^{-1}⁡(\frac{-1}{2})\) ?
a) –\(\frac{π}{2}\)
b) π
c) \(\frac{π}{2}\)
d) \(\frac{17π}{6}\)
View Answer

Answer: c
Explanation: \(cos^{-1}⁡(\frac{⁡1}{2})=\frac{π}{3} and sin^{-1}⁡(-\frac{⁡1}{2})=-\frac{π}{6}\)
∴ 5 \(cos^{-1}⁡\frac{1}{2}+7 sin^{-1}⁡(-\frac{1}{2}) =5(\frac{π}{3})+7(-\frac{π}{6})\)
=\(\frac{5π}{3}-\frac{7π}{6}=\frac{10π-7π}{6}=\frac{3π}{6}=\frac{π}{2}\)

7. Find the value of \(sin^{-1}⁡(sin⁡ \frac{4π}{3})\) is _______
a) π
b) \(\frac{π}{3}\)
c) \(\frac{4π}{3}\)
d) –\(\frac{π}{3}\)
View Answer

Answer: d
Explanation: \(sin^{-1}⁡(sin⁡x)\)=x, x∈\([-\frac{π}{2},\frac{π}{2}]\)
∴\(sin^{-1} (sin⁡ \frac{4π}{3})=sin^{-1}⁡(sin⁡(π+\frac{π}{3}))=sin^{-1}⁡(sin⁡(\frac{-π}{3}))= -\frac{π}{3}\).

8. Find the value of \(cos⁡(sin^{-1}⁡\frac{\sqrt{3}}{2})\) is _____
a) \(\frac{\sqrt{3}}{2}\)
b) \(\frac{1}{4}\)
c) \(\frac{1}{2}\)
d) 0
View Answer

Answer: c
Explanation: \(sin⁡\frac{π}{3}=\frac{\sqrt{3}}{2}\)
∴\(sin^{-1}⁡\frac{\sqrt{3}}{2}=\frac{π}{3}\)
⇒\(cos⁡(sin^{-1}⁡\frac{\sqrt{3}}{2})=cos⁡(\frac{π}{3})=\frac{1}{2}\).

9. If \(cos^{-1}⁡x=y\), then which of the following is correct?
a) 0 ≤ y ≤ π
b) 0 < y < π
c) –\(\frac{π}{2}≤y≤\frac{π}{2}\)
d) –\(\frac{π}{2}<y<\frac{π}{2}\)
View Answer

Answer: a
Explanation: Given that, \(cos^{-1}⁡x=y\)
The range of principle values for the inverse trigonometric function \(cos^{-1}\) is [0,π].
Hence, 0≤y≤π.

10. \(sin^{-1}⁡x\) is same as \((sin⁡x)^{-1}\).
a) True
b) False
View Answer

Answer: b
Explanation: The given statement is false. \(sin^{-1}⁡x\) is not same as \((sin⁡x)^{-1}\). \(sin^{-1}⁡x\) is an inverse trigonometric function whereas \((sin⁡x)^{-1}\) is just the reciprocal of sin⁡x i.e. \(sin⁡x=\frac{1}{sin⁡x}\).

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