# Class 12 Maths MCQ – Inverse Trigonometric Functions

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) $$\frac{π}{4}$$

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) $$\frac{π}{2}$$

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

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

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]

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) $$\frac{17π}{6}$$

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) –$$\frac{π}{3}$$

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

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) –$$\frac{π}{2}<y<\frac{π}{2}$$

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

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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