Inverse Trigonometric Functions MCQ with Solutions

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Inverse Trigonometric Functions MCQ with answers and solutions from NCERT Class 12 Maths. Revise all 7 important questions step by step with clear explanations. You can also watch the video solutions for better understanding. This is in continuation of the previous topic on Relations and Functions MCQs.

Ready? Okay, let’s begin with the first question!

Inverse Trigonometric Functions MCQ#1 (NCERT Exercise 2.1 – Question 13)

📍If $\sin^{-1} x = y$ , then

Options:

(A) $0 \le y \le \pi$
(B) $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
(C) $0 < y < \pi$
(D) $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$

Answer:
✅ Correct option: (B)

Explanation:
By definition, the principal value branch of $\sin^{-1} x$ is chosen so that
$-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$ .

This ensures $\sin y = x$ gives a unique value of $y$ for each $x \in [-1,1]$ .
Hence, $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$ .

Watch Video Solution on YouTube

Let me tell you all of these 7 Inverse Trigonometric Functions MCQ are very simple and easy to understand and therefore scoring too. But you must practice them regularly.

So, my suggestion to you is to solve these NCERT Inverse Trigonometric Functions MCQ a good number of times and feel confident. And if you ever need any clarification, just drop a comment — I’ll be more than happy to help.

Inverse Trigonometric Functions MCQ#2 (NCERT Exercise 2.1 – Question 14)

📍 $\tan^{-1} \sqrt{3} - \sec^{-1} (-2)$ is equal to

Options:

(A) $\pi$
(B) $-\dfrac{\pi}{3}$
(C) $\dfrac{\pi}{3}$
(D) $\dfrac{2\pi}{3}$

Answer:
✅ Correct Option: (B)

Explanation:

We know,
$\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$

Now,
$\sec^{-1}(-2)$ means we have to find an angle $\theta$ such that
$\sec\theta = -2$

✅️ This is possible when $\cos\theta = -\dfrac{1}{2}$

The principal value of $\sec^{-1}x$ lies in $[0,\pi] - {\dfrac{\pi}{2}}$

So, $\theta = \dfrac{2\pi}{3}$

Hence,
$\sec^{-1}(-2) = \dfrac{2\pi}{3}$

Now,
$\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}$

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Inverse Trigonometric Function MCQ#3 (NCERT Exercise 2.2 – Question 13)

📍 $\cos^{-1}\big(\cos \tfrac{7\pi}{6}\big)$ is equal to

Options:

(A) $\tfrac{7\pi}{6}$
(B) $\tfrac{5\pi}{6}$
(C) $\tfrac{\pi}{3}$
(D) $\tfrac{\pi}{6}$

Answer:
✅ Correct Option: (B)

Explanation:

$\cos\frac{7\pi}{6} = \cos\big(\pi + \tfrac{\pi}{6}\big) = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ .

We need $\theta \in [0,\pi]$ with $\cos\theta = -\frac{\sqrt{3}}{2}$ = – $\cos\frac{\pi}{6} = \cos\big(\pi - \tfrac{\pi}{6}\big)$

That is $\theta = \frac{5\pi}{6}$ .

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Did you know, there are about 100 MCQs in NCERT textbooks for Class 12 (both Part 1 and Part 2)? I shall be taking one Chapter at a time and show you the explanations for each of the questions.

Moreover, I have also covered these solutions in the video form and they are freely available on my YouTube channel, @Mathsbetter. And here also, in each of the 7 Inverse Trigonometric Functions MCQ on this page, I have provided the direct links to each and every MCQ for you to understand the solutions in a very simple and easy manner.

Inverse Trigonometric Functions MCQ#4 (NCERT Exercise 2.2 – Question 14)

📍 $\sin\Big(\tfrac{\pi}{3} - \sin^{-1}\big(-\tfrac{1}{2}\big)\Big)$ is equal to

Options:

(A) $\tfrac{1}{2}$
(B) $\tfrac{1}{3}$
(C) $\tfrac{1}{4}$
(D) $1$

Answer:
✅ Correct Option: (D)

Explanation:

Let $\alpha = \sin^{-1}\big(-\tfrac{1}{2}\big)$ . Principal value: $\alpha = -\tfrac{\pi}{6}$ (since $\sin(-\tfrac{\pi}{6}) = -\tfrac{1}{2}$ ).

So the angle becomes $\tfrac{\pi}{3} - \alpha = \tfrac{\pi}{3} - \big(-\tfrac{\pi}{6}\big) = \tfrac{\pi}{3} + \tfrac{\pi}{6} = \tfrac{\pi}{2}$ .

$\sin\big(\tfrac{\pi}{2}\big) = 1$ .

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Inverse Trigonometric Functions MCQ#5 (NCERT Exercise 2.2 – Question 15)

📍 $\tan^{-1}\sqrt{3} - \cot^{-1}\big(-\sqrt{3}\big)$ is equal to

Options:

(A) $\pi$
(B) $-\tfrac{\pi}{2}$
(C) $0$
(D) $2\sqrt{3}$

Answer:
✅ Correct Option: (B)

Explanation:

$\tan^{-1}\sqrt{3} = \tfrac{\pi}{3}$ (since $\tan\tfrac{\pi}{3}=\sqrt{3}$ ).

For $\cot^{-1}(-\sqrt{3})$ , find $\theta\in(0,\pi)$ with $\cot\theta = -\sqrt{3}$ .
→ $\tan\theta = -\tfrac{1}{\sqrt{3}}$ , reference angle $\tfrac{\pi}{6}$ , negative tangent in its PVB occurs in Quadrant II, so $\theta = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$ .

Therefore $\cot^{-1}(-\sqrt{3}) = \tfrac{5\pi}{6}$ .

Now compute: $\tfrac{\pi}{3} - \tfrac{5\pi}{6} = \tfrac{2\pi}{6} - \tfrac{5\pi}{6} = -\tfrac{3\pi}{6} = -\tfrac{\pi}{2}$ .

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Inverse Trigonometric Function MCQ#6 (NCERT Miscellaneous Exercise Ch. 2 – Question 13)

📍 $\sin(\tan^{-1}x),\ |x|<1$ is equal to

Options:

(A) $\dfrac{x}{\sqrt{1 - x^{2}}}$
(B) $\dfrac{1}{\sqrt{1 - x^{2}}}$
(C) $\dfrac{1}{\sqrt{1 + x^{2}}}$
(D) $\dfrac{x}{\sqrt{1 + x^{2}}}$

Answer:
✅ Correct Option: (D)

Explanation:

Let $\theta = \tan^{-1}x$ .
Then $\tan\theta = x = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\text{P}}{\text{B}} = \dfrac{x}{1}$ .

So, hypotenuse $= \sqrt{1 + x^{2}}$ .

Hence,
$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{\text{P}}{\text{H}} = \dfrac{x}{\sqrt{1 + x^{2}}}$ .

Therefore,
$\sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1 + x^{2}}}$ .

Watch Video Solution on YouTube

If you have been following this website as a resource for some of the important questions. For example, how to Integrate Square Root of tan x, or may be a complete guide to look at using Matrix Method in solving a system of linear equations.

Trust me, similarly, this series on NCERT Class 12 MCQs is going to help you in many ways.

Moving onto the next Inverse Trigonometric Functions MCQ in this part of the series.

Inverse Trigonometric Functions MCQ#7 (NCERT Miscellaneous Exercise Ch. 2 – Question 14)

📍 $\sin^{-1}(1 - x) - 2\sin^{-1}x = \dfrac{\pi}{2}$ , then $x$ is equal to

Options:

(A) $0,\ \dfrac{1}{2}$
(B) $1,\ \dfrac{1}{2}$
(C) $0$
(D) $\dfrac{1}{2}$

Answer:
✅ Correct option: (C)

Explanation:

Let $\sin^{-1}x=\theta$ . Then $x=\sin\theta$ and the given equation becomes

$\sin^{-1}(1-\sin\theta) - 2\theta = \dfrac{\pi}{2}$ .

So, we get,

$1-\sin\theta = \sin\Big(\dfrac{\pi}{2}+2\theta\Big) = \cos(2\theta) = 1 - 2\sin^{2}\theta.$

So

$1-\sin\theta = 1 - 2\sin^{2}\theta \Rightarrow 2\sin^{2}\theta - \sin\theta = 0 \Rightarrow \sin\theta(2\sin\theta - 1)=0.$

Hence the algebraic candidates are

$\sin\theta = 0 \Rightarrow x=0,$
or
$\sin\theta = \dfrac{1}{2} \Rightarrow x=\dfrac{1}{2}.$ 👉 and for $x = 0:$
$\sin^{-1}(1 - 0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \dfrac{\pi}{2}$ ✅ Satisfied.

👉 For $x = \dfrac{1}{2}:$
$\sin^{-1}(1 - \tfrac{1}{2}) - 2\sin^{-1}(\tfrac{1}{2}) = \sin^{-1}(\tfrac{1}{2}) - 2(\tfrac{\pi}{6}) = \tfrac{\pi}{6} - \tfrac{\pi}{3} = -\tfrac{\pi}{6}$ ❌ Not satisfied.

So only x = 0 satisfies the given equation.

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Quick Answer Key

Q1 → (B)
Q2 → (B)
Q3 → (B)
Q4 → (D)
Q5 → (B)
Q6 → (D)
Q7 → (C)

Closing Note

That completes all the important NCERT Class 12 Maths Inverse Trigonometric Functions MCQ from chapter 2.
I hope the explanations and video solutions made things clearer and gave you more confidence. Stay tuned for the next chapter, where we’ll cover all the important Application of Derivatives MCQ (Chapter 6 from NCERT) with the same clarity and video support.

👉 Make sure to practice these questions again to strengthen your concepts.
👉 Watch the video solutions for a clearer and faster revision.

Keep practicing, and you’ll master inverse trigonometric functions easily! 🚀

Inverse Trigonometric Functions MCQ with Solutions

Inverse Trigonometric Functions: Chapter 2 Links

Inverse Trigonometric Functions MCQ#1 (NCERT Exercise 2.1 – Question 13)

Inverse Trigonometric Functions MCQ#2 (NCERT Exercise 2.1 – Question 14)

Inverse Trigonometric Function MCQ#3 (NCERT Exercise 2.2 – Question 13)

Inverse Trigonometric Functions MCQ#4 (NCERT Exercise 2.2 – Question 14)

Inverse Trigonometric Functions MCQ#5 (NCERT Exercise 2.2 – Question 15)

Inverse Trigonometric Function MCQ#6 (NCERT Miscellaneous Exercise Ch. 2 – Question 13)

Inverse Trigonometric Functions MCQ#7 (NCERT Miscellaneous Exercise Ch. 2 – Question 14)

Quick Answer Key

Closing Note

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Inverse Trigonometric Functions: Chapter 2 Links

Inverse Trigonometric Functions MCQ#1 (NCERT Exercise 2.1 – Question 13)

Inverse Trigonometric Functions MCQ#2 (NCERT Exercise 2.1 – Question 14)

Inverse Trigonometric Function MCQ#3 (NCERT Exercise 2.2 – Question 13)

Inverse Trigonometric Functions MCQ#4 (NCERT Exercise 2.2 – Question 14)

Inverse Trigonometric Functions MCQ#5 (NCERT Exercise 2.2 – Question 15)

Inverse Trigonometric Function MCQ#6 (NCERT Miscellaneous Exercise Ch. 2 – Question 13)

Inverse Trigonometric Functions MCQ#7 (NCERT Miscellaneous Exercise Ch. 2 – Question 14)

Quick Answer Key

Closing Note

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