CBSE Class 12 Mathematics Inverse Trigonometric Functions VBQs Set C

Read and download the CBSE Class 12 Mathematics Inverse Trigonometric Functions VBQs Set C. Designed for the 2025-26 academic year, these Value Based Questions (VBQs) are important for Class 12 Mathematics students to understand moral reasoning and life skills. Our expert teachers have created these chapter-wise resources to align with the latest CBSE, NCERT, and KVS examination patterns.

VBQ for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

For Class 12 students, Value Based Questions for Chapter 2 Inverse Trigonometric Functions help to apply textbook concepts to real-world application. These competency-based questions with detailed answers help in scoring high marks in Class 12 while building a strong ethical foundation.

Chapter 2 Inverse Trigonometric Functions Class 12 Mathematics VBQ Questions with Answers

Question. Fill in the blanks.
(i) The principal value of \( \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \) is _____________ .
(ii) The set of values of \( \sec^{-1} \left( \frac{1}{2} \right) \) is _____________ .
(iii) The value of \( \cos(\sin^{-1} x + \cos^{-1} x), |x| \le 1 \) is _____________ .
(iv) The value of \( \sin^{-1} \left( \cos \left( \frac{43\pi}{5} \right) \right) \) is _____________ .
Answer: (i) \( \frac{5\pi}{6} \), (ii) \( \phi \), (iii) \( 0 \), (iv) \( -\frac{\pi}{10} \)

Very Short Answer Questions:

Question. What is the domain of the function \( \sin^{-1} x \)?
Answer: \( [-1, 1] \)

Question. Write the principal value of \( \cot^{-1}(-\sqrt{3}) \).
Answer: \( \frac{5\pi}{6} \)

Question. If \( 4 \cos^{-1} x + \sin^{-1} x = \pi \), then find the value of \( x \).
Answer: \( x = \frac{\sqrt{3}}{2} \)

Question. Evaluate: \( \tan (\tan^{-1} (-4)) \)
Answer: \( -4 \)

Question. Write the principal value of \( \cos^{-1} \left( \frac{1}{2} \right) - 2 \sin^{-1} \left( -\frac{1}{2} \right) \).
Answer: \( \frac{2\pi}{3} \)

Question. Write the value of \( \sin(\cot^{-1} x) \).
Answer: \( \frac{1}{\sqrt{1 + x^2}} \)

Short Answer Questions-I:

Question. Find the value of \( \sin(2 \sin^{-1} (0.6)) \).
Answer: \( 0.96 \)

Question. Show that \( \sin^{-1}(2x\sqrt{1-x^2}) = 2 \sin^{-1} x, -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \).
Answer: Let \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \).
LHS \( = \sin^{-1}(2\sin \theta \sqrt{1 - \sin^2 \theta}) \)
\( = \sin^{-1}(2\sin \theta \cos \theta) \)
\( = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \sin^{-1} x \).
Hence Proved.

Question. Write the simplest form of \( \tan^{-1} \frac{1}{\sqrt{x^2 - 1}}, |x| > 1 \).
Answer: \( \frac{\pi}{2} - \sec^{-1} x \)

Question. Prove that: \( 3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), x \in \left[ -\frac{1}{2}, \frac{1}{2} \right] \).
Answer: Let \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \).
RHS \( = \sin^{-1}(3\sin \theta - 4\sin^3 \theta) \)
\( = \sin^{-1}(\sin 3\theta) = 3\theta = 3 \sin^{-1} x \).
Hence Proved.

Question. Write the simplest form of \( \tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right), |x| < a \).
Answer: \( \sin^{-1} \frac{x}{a} \)

Question. Write the principal value of \( \tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}) \).
Answer: \( -\frac{\pi}{2} \)

Short Answer Questions-II:

Question. Prove that: \( \cos \left( \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \right) = \frac{6}{5\sqrt{13}} \).
Answer: Let \( \sin^{-1} \frac{3}{5} = A \Rightarrow \sin A = \frac{3}{5}, \cos A = \frac{4}{5} \).
Let \( \cot^{-1} \frac{3}{2} = B \Rightarrow \cot B = \frac{3}{2} \Rightarrow \sin B = \frac{2}{\sqrt{13}}, \cos B = \frac{3}{\sqrt{13}} \).
\( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
\( = \frac{4}{5} \cdot \frac{3}{\sqrt{13}} - \frac{3}{5} \cdot \frac{2}{\sqrt{13}} = \frac{12 - 6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}} \).
Hence Proved.

Question. Solve: \( \tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1} \frac{8}{31} \).
Answer: \( x = \frac{1}{4} \)

Question. If \( \sin[\cot^{-1}(x + 1)] = \cos(\tan^{-1} x) \), then find \( x \).
Answer: \( x = -\frac{1}{2} \)

Question. Evaluate: \( \tan \left\{ 2 \tan^{-1} \left( \frac{1}{5} \right) + \frac{\pi}{4} \right\} \).
Answer: \( \frac{17}{7} \)

Question. Prove the following:
\( \cot^{-1} \left( \frac{xy + 1}{x - y} \right) + \cot^{-1} \left( \frac{yz + 1}{y - z} \right) + \cot^{-1} \left( \frac{zx + 1}{z - x} \right) = 0 \) (\( 0 < xy, yz, zx < 1 \)).

Answer: RHS \( = (\tan^{-1} x - \tan^{-1} y) + (\tan^{-1} y - \tan^{-1} z) + (\tan^{-1} z - \tan^{-1} x) = 0 \).
Hence Proved.

Question. Prove the following:
\( \sin \left[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right] = 1, 0 < x < 1 \).

Answer: Let \( x = \tan \theta \).
\( = \sin[\tan^{-1}(\cot 2\theta) + \cos^{-1}(\cos 2\theta)] \)
\( = \sin[\tan^{-1}(\tan(\frac{\pi}{2} - 2\theta)) + 2\theta] \)
\( = \sin[\frac{\pi}{2} - 2\theta + 2\theta] = \sin \frac{\pi}{2} = 1 \).
Hence Proved.

Question. Prove that: \( 2 \tan^{-1} \left( \sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2} \right) = \cos^{-1} \left( \frac{a \cos x + b}{a + b \cos x} \right) \).
Answer: Standard transformation of inverse trigonometric functions. Hence Proved.

Question. If \( \tan^{-1} \left( \frac{1}{1 + 1.2} \right) + \tan^{-1} \left( \frac{1}{1 + 2.3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n(n + 1)} \right) = \tan^{-1} \theta \), then find the value of \( \theta \).
Answer: \( \theta = \frac{n}{n + 2} \)

Question. Prove that: \( \tan^{-1} \left( \frac{\sqrt{1 + \cos x} + \sqrt{1 - \cos x}}{\sqrt{1 + \cos x} - \sqrt{1 - \cos x}} \right) = \frac{\pi}{4} + \frac{x}{2} \), where \( \pi < x < \frac{3\pi}{2} \).
Answer: By substituting \( \cos x = 2 \cos^2(x/2) - 1 \) and \( \cos x = 1 - 2 \sin^2(x/2) \). Hence Proved.

Question. Solve for \( x \): \( \tan^{-1} \left( \frac{2 - x}{2 + x} \right) = \frac{1}{2} \tan^{-1} \frac{x}{2}, x > 0 \).
Answer: \( x = \frac{2}{\sqrt{3}} \)

Question. Prove that: \( 2 \sin^{-1} \left( \frac{3}{5} \right) - \tan^{-1} \left( \frac{17}{31} \right) = \frac{\pi}{4} \).
Answer: Using \( 2 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7} \) and then \( \tan^{-1} \frac{24}{7} - \tan^{-1} \frac{17}{31} = \frac{\pi}{4} \). Hence Proved.

Question. Prove that \( \tan^{-1} \left( \frac{6x - 8x^3}{1 - 12x^2} \right) - \tan^{-1} \left( \frac{4x}{1 - 4x^2} \right) = \tan^{-1} 2x \); \( |2x| < \frac{1}{\sqrt{3}} \).
Answer: Let \( 2x = \tan \theta \). Then use the formula for \( \tan 3\theta \) and \( \tan 2\theta \). Hence Proved.

Question. Solve for \( x \): \( \tan^{-1} \left[ \frac{x - 3}{x - 4} \right] + \tan^{-1} \left[ \frac{x + 3}{x + 4} \right] = \frac{\pi}{4} \).
Answer: \( x = \pm \sqrt{\frac{17}{2}} \)

Question. If \( \tan^{-1} x - \cot^{-1} x = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right), x > 0 \), find the value of \( x \) and hence find the value of \( \sec^{-1} \left( \frac{2}{x} \right) \).
Answer: \( x = \sqrt{3} \); \( \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6} \)

Question. If \( \sin^{-1} \left( \frac{3}{x} \right) + \sin^{-1} \left( \frac{4}{x} \right) = \frac{\pi}{2} \), then find the value of \( x \).
Answer: \( x = 5 \)

Question. Find the value of \( x \), if \( \tan[\sec^{-1}(1/x)] = \sin(\tan^{-1} 2), x > 0 \).
Answer: \( x = \frac{\sqrt{5}}{3} \)

Question. Prove that \( \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \sin^{-1} \left( \frac{4}{5} \right) \).
Answer: \( \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \tan^{-1} \left( \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{2}{36}} \right) = \tan^{-1} \frac{1}{2} = \frac{1}{2} (2 \tan^{-1} \frac{1}{2}) = \frac{1}{2} \sin^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 + \frac{1}{4}} \right) = \frac{1}{2} \sin^{-1} \frac{4}{5} \). Hence Proved.

VBQs for Chapter 2 Inverse Trigonometric Functions Class 12 Mathematics

Students can now access the Value-Based Questions (VBQs) for Chapter 2 Inverse Trigonometric Functions as per the latest CBSE syllabus. These questions have been designed to help Class 12 students understand the moral and practical lessons of the chapter. You should practicing these solved answers to improve improve your analytical skills and get more marks in your Mathematics school exams.

Expert-Approved Chapter 2 Inverse Trigonometric Functions Value-Based Questions & Answers

Our teachers have followed the NCERT book for Class 12 Mathematics to create these important solved questions. After solving the exercises given above, you should also refer to our NCERT solutions for Class 12 Mathematics and read the answers prepared by our teachers.

Improve your Mathematics Scores

Daily practice of these Class 12 Mathematics value-based problems will make your concepts better and to help you further we have provided more study materials for Chapter 2 Inverse Trigonometric Functions on studiestoday.com. By learning these ethical and value driven topics you will easily get better marks and also also understand the real-life application of Mathematics.

Where can I find 2025-26 CBSE Value Based Questions (VBQs) for Class 12 Mathematics Chapter Chapter 2 Inverse Trigonometric Functions?

The latest collection of Value Based Questions for Class 12 Mathematics Chapter Chapter 2 Inverse Trigonometric Functions is available for free on StudiesToday.com. These questions are as per 2026 academic session to help students develop analytical and ethical reasoning skills.

Are answers provided for Class 12 Mathematics Chapter Chapter 2 Inverse Trigonometric Functions VBQs?

Yes, all our Mathematics VBQs for Chapter Chapter 2 Inverse Trigonometric Functions come with detailed model answers which help students to integrate factual knowledge with value-based insights to get high marks.

What is the importance of solving VBQs for Class 12 Chapter Chapter 2 Inverse Trigonometric Functions Mathematics?

VBQs are important as they test student's ability to relate Mathematics concepts to real-life situations. For Chapter Chapter 2 Inverse Trigonometric Functions these questions are as per the latest competency-based education goals.

How many marks are usually allocated to VBQs in the CBSE Mathematics paper?

In the current CBSE pattern for Class 12 Mathematics, Chapter 2 Inverse Trigonometric Functions Value Based or Case-Based questions typically carry 3 to 5 marks.

Can I download Mathematics Chapter Chapter 2 Inverse Trigonometric Functions VBQs in PDF for free?

Yes, you can download Class 12 Mathematics Chapter Chapter 2 Inverse Trigonometric Functions VBQs in a mobile-friendly PDF format for free.