NCERT Solutions for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

NCERT Solutions for Class 12 Mathematics Chapter 2 - Inverse Trigonometric Functions

Welcome to the Chapter 2 - Inverse Trigonometric Functions, Class 12 Mathematics NCERT Solutions page. Here, we provide detailed question answers for Chapter 2 - Inverse Trigonometric Functions. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Inverse Trigonometric Functions and excel in their exams. By going through these Inverse Trigonometric Functions question answers, you can strengthen your foundation and improve your performance in Class 12 Mathematics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

Q 1:

Find the principal value of \begin{align} sin^{-1}\left(-\frac{1}{2}\right)\end{align}

\begin{align} sin^{-1}\left(-\frac{1}{2}\right)=y \;\;Then\;\; sin y = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align}

We know that the range of the principal value branch of sin⁻¹ is

\begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \;\;sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\end{align}

Therefore, the principal value of

\begin{align} sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}\end{align}

Q 10:

Find the principal value of \begin{align} cosec^{-1}\left({-\sqrt2}\right)\end{align}

\begin{align} Let \;\; cosec^{-1}\left({-\sqrt2}\right)=y \;\;Then\;\; cosec y = -{\sqrt2} =- cosec\left(\frac{\pi}{4}\right) = cosec\left(-\frac{\pi}{4}\right)\end{align}

We know that the range of the principal value branch of cosec⁻¹ is

\begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \left(0 \right) and \;\;cosec\left(-\frac{\pi}{4}\right) = -\sqrt2.\end{align}

Therefore, the principal value of

\begin{align} cosec^{-1}\left(-\sqrt2\right) is -\frac{\pi}{4}\end{align}

Q 11:

Find the principal value of \begin{align} tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}

\begin{align} Let \;\; tan^{-1}(1)=x. \;\;Then\;\; tan x = 1 = tan\left(\frac{\pi}{4}\right).\end{align}

\begin{align} \therefore tan^{-1}(1)=tan\left(\frac{\pi}{4}\right)\end{align}

\begin{align} Let \;\;cos^{-1}\left(-\frac{1}{2}\right)=y. \;\;Then,\;\; cos y = -\frac{1}{2} = -cos\left(\frac{\pi}{3}\right)= cos\left(\pi - \frac{\pi}{3}\right) = cos\left(\frac{2\pi}{3}\right)\end{align}

\begin{align} \therefore cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\end{align}

\begin{align} Let \;\; sin^{-1}\left(-\frac{1}{2}\right)=z. \;\;Then,\;\; sin z = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align}

\begin{align} \therefore sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\end{align}

\begin{align} \therefore tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}

\begin{align} =\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}\end{align}

\begin{align} =\frac{3\pi + 8\pi -2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}\end{align}

Q 12:

Find the principal value of \begin{align} cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}

\begin{align} Let \;\;cos^{-1}\left(\frac{1}{2}\right) =x. \;\;Then,\;\; cos x = \frac{1}{2} = cos\left(\frac{\pi}{3}\right).\end{align}

\begin{align} \therefore cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\end{align}

\begin{align} Let \;\; sin^{-1}\left(\frac{1}{2}\right)=y. \;\;Then,\;\; sin y = \frac{1}{2} = sin\left(\frac{\pi}{6}\right).\end{align}

\begin{align} \therefore sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\end{align}

\begin{align} \therefore cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}

\begin{align} =\frac{\pi}{3} + \frac{2\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\end{align}

Q 13:

Find the value of if \begin{align} sin^{-1} x= y,\end{align}
then
\begin{align} (A) 0 ≤ y ≤ x \;\;\;\;\; (B) -\frac{\pi}{2} ≤ y ≤ \frac{\pi}{2} \end{align}
\begin{align} (C)\; 0 < y < \pi \;\;\;\;\; (D) -\frac{\pi}{2} < y < \frac{\pi}{2} \end{align}

It is given that sin⁻¹ x = y.

We know that the range of the principal value branch of sin⁻¹ is

\begin{align} \left[-\frac{\pi}{2} ,\frac{\pi}{2}\right] \end{align}

Therefore,

\begin{align} (B) -\frac{\pi}{2} ≤ y ≤ \frac{\pi}{2} \end{align}

Q 14:

Find the value of if \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}
is equal to
\begin{align} (A) \pi \;\;(B) -\frac{\pi}{3}\;\; (C) \frac{\pi}{3} \;\;(D) \frac{2\pi}{3}\end{align}

\begin{align} Let \;\; tan^{-1}(\sqrt 3)=x. \;\;Then\;\; tan x = \sqrt 3 = tan\left(\frac{\pi}{3}\right).\end{align}

We know that the range of the principal value branch of tan⁻¹ is

\begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\end{align}

\begin{align} \therefore tan^{-1}\sqrt3=\frac{\pi}{3}\end{align}

\begin{align} Let \;\; sec^{-1}\left(-2\right)=y \;\;Then\;\; sec y = -2 = -sec\left(\frac{\pi}{3}\right) = sec\left(\pi - \frac{\pi}{3}\right) = sec\left(\frac{2\pi}{3}\right)\end{align}

We know that the range of the principal value branch of sec⁻¹ is

\begin{align} \left[0,\pi\right] - \left(\frac{\pi}{2}\right)\end{align}

\begin{align} \therefore sec^{-1}({-2})=\frac{2\pi}{3}\end{align}

Hence, \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}

\begin{align} =\frac{\pi}{3} - \frac{2\pi}{3}=-\frac{\pi}{3}\end{align}

Q 2:

Find the principal value of \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right)\end{align}

\begin{align} Let\;\; cos^{-1}\left(\frac{\sqrt3}{2}\right)=y, \;\;Then,\;\; cos y = \frac{\sqrt3}{2} = cos\left(\frac{\pi}{6}\right)\end{align}

We know that the range of the principal value branch of cos⁻¹ is

\begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{\pi}{6}\right) = \frac{\sqrt3}{2}\end{align}

Therefore, the principal value of

\begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right) is \frac{\pi}{6}\end{align}

Q 3:

Find the principal value of \begin{align} cosec^{-1}\left({2}\right)\end{align}

\begin{align} Let \;\; cosec^{-1}\left({2}\right)=y \;\;Then\;\; cosec y = 2 = cosec\left(\frac{\pi}{6}\right)\end{align}

We know that the range of the principal value branch of cosec−1 is

\begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \left(0 \right). \end{align}

Therefore, the principal value of

\begin{align} \frac{\pi}{6}\end{align}

Q 4:

Find the principal value of \begin{align} tan^{-1}\left(-\sqrt3\right)\end{align}

\begin{align} Let \;\; tan^{-1}\left(-\sqrt3\right)=y \;\;Then\;\; tan y = -\sqrt3 = -tan\left(\frac{\pi}{3}\right)= tan\left(-\frac{\pi}{3}\right)\end{align}

We know that the range of the principal value branch of tan⁻¹ is

\begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right) and \;\;tan\left(-\frac{\pi}{3}\right) = -\sqrt3\end{align}

Therefore, the principal value of

\begin{align} tan^{-1}\left(-\sqrt 3\right) is -\frac{\pi}{3}\end{align}

Q 5:

Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{2}\right)\end{align}

\begin{align} Let\;\; cos^{-1}\left(-\frac{1}{2}\right)=y, \;\;Then,\;\; cos y = -\frac{1}{2} = - cos\left(\frac{\pi}{3}\right)=cos\left(\pi - \frac{\pi}{3}\right) = cos\left(\frac{2\pi}{3}\right)\end{align}

We know that the range of the principal value branch of cos⁻¹ is

\begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\end{align}

Therefore, the principal value of

\begin{align} cos^{-1}\left(-\frac{1}{2}\right) is \frac{2\pi}{3}\end{align}

Q 6:

Find the principal value of \begin{align} tan^{-1}\left(-1\right)\end{align}

\begin{align} Let \;\; tan^{-1}\left(-1\right)=y. \;\;Then,\;\; tan y = -1 = -tan\left(\frac{\pi}{4}\right)= tan\left(-\frac{\pi}{4}\right)\end{align}

We know that the range of the principal value branch of tan−1 is

\begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right) and \;\;tan\left(-\frac{\pi}{4}\right) = - 1\end{align}

Therefore, the principal value of

\begin{align} tan^{-1}\left(- 1\right) is -\frac{\pi}{4}\end{align}

Q 7:

Find the principal value of \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right)\end{align}

\begin{align} Let \;\; sec^{-1}\left(\frac{2}{\sqrt3}\right)=y \;\;Then\;\; sec y = \frac{2}{\sqrt3} = sec\left(\frac{\pi}{6}\right)\end{align}

We know that the range of the principal value branch of sec⁻¹ is

\begin{align} \left[0,\pi\right] - \left(\frac{\pi}{2}\right) and \;\;sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt3}\end{align}

Therefore, the principal value of

\begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right) is \frac{\pi}{6}\end{align}

Q 8:

Find the principal value of \begin{align} cot^{-1}\left(\sqrt3\right)\end{align}

\begin{align} Let \;\; cot^{-1}\left(\sqrt3\right)=y. \;\;Then,\;\; cot y = \sqrt3 = cot\left(\frac{\pi}{6}\right).\end{align}

We know that the range of the principal value branch of cot⁻¹ is

\begin{align} \left(0, \pi\right) and \;\;cot\left(\frac{\pi}{6}\right) = \sqrt3.\end{align}

Therefore, the principal value of

\begin{align} cot^{-1}\left(\sqrt 3\right) is \frac{\pi}{6}\end{align}

Q 9:

Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right)\end{align}

\begin{align} Let\;\; cos^{-1}\left(-\frac{1}{\sqrt2}\right)=y, \;\;Then,\;\; cos y = -\frac{1}{\sqrt2} = - cos\left(\frac{\pi}{4}\right)=cos\left(\pi - \frac{\pi}{4}\right) = cos\left(\frac{3\pi}{4}\right)\end{align}

We know that the range of the principal value branch of cos⁻¹ is

\begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt2}\end{align}

Therefore, the principal value of

\begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right) is \frac{3\pi}{4}\end{align}

Frequently Asked Questions about Inverse Trigonometric Functions - Class 12 Mathematics

- 1. How many questions are covered in Inverse Trigonometric Functions solutions?
- All questions from Inverse Trigonometric Functions are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
- 2. Are the solutions for Inverse Trigonometric Functions helpful for exam preparation?
- Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
- 3. Can I find solutions to all exercises in Inverse Trigonometric Functions?
- Yes, we provide solutions to all exercises, examples, and additional questions from Inverse Trigonometric Functions with detailed explanations.
- 4. How do these solutions help in understanding Inverse Trigonometric Functions concepts?
- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
- 5. Are there any tips for studying Inverse Trigonometric Functions effectively?
- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Inverse Trigonometric Functions

The Inverse Trigonometric Functions is an important chapter of 12 Mathematics. This chapter’s important topics like Inverse Trigonometric Functions are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

Latest Blog Posts

Stay updated with our latest educational content and study tips

Smart Questions to Ask in a Parent-Teacher Meeting | PTM Made Easy

Parent-Teacher Meetings (PTMs) are more than quick updates on marks — they’re a chance to build a real partnership between home and school. A good Parent-Teacher Meeting conversation helps parents see beyond grades. It opens up insights about a child’s strengths, struggles, emotions and even hidden talents. When parents participate actively, they don’t just track […]

The Secret to Smarter Learning — Building Strong Critical Thinking Skills

In today’s world of endless information , knowing how to think is more important than knowing what to think . From school projects to real – life decisions , critical thinking helps students question ideas , analyze facts and form logical conclusions . But what exactly does critical thinking mean ? Simply put , it’s […]

Study Smarter, Not Harder: Build Productive Habits That Stick

Every student dreams of better grades , stronger focus and more study time – but the real challenge isn’t starting, it’s staying consistent . Building productive study habits is not about studying all day , it’s about studying smart . In today’s fast – paced digital world, distractions are everywhere – from endless phone notifications […]

The Hidden Risks of Online Gaming for Children — Is your child safe while gaming online?

Online gaming has rapidly become one of the most popular pastimes among children. Whether it’s multiplayer mobile games , PC adventures or console challenges , kids are spending more time than ever in the virtual world . On the surface, gaming seems entertaining and even educational – improving hand- eye coordination , teamwork and problem […]

View All Blog Posts

Benefits of Using Our NCERT Solutions for Class

When it comes to excelling in your studies, having a well-structured study guide can make a huge difference. Our NCERT Solutions for Class provide you with a comprehensive, easy-to-understand, and exam-focused resource that is specifically tailored to help you maximize your potential. Here are some of the key benefits of using our NCERT solutions for effective learning and high scores:

NCERT Solutions for Effective Exam Preparation

Preparing for exams requires more than just reading through textbooks. It demands a structured approach to understanding concepts, solving problems, and revising thoroughly. Here’s how our NCERT solutions can enhance your exam preparation:

Clear Understanding of Concepts: Our NCERT solutions are designed to break down complex topics into simple, understandable language, making it easier for students to grasp essential concepts in . This helps in building a solid foundation for each chapter, which is crucial for scoring high marks.
Step-by-Step Solutions: Each solution is presented in a detailed, step-by-step manner. This approach not only helps you understand how to reach the answer but also equips you with the right techniques to tackle similar questions in exams.
Access to Important Questions: We provide a curated list of important questions and commonly asked questions in exams. By practicing these questions, you can familiarize yourself with the types of problems that are likely to appear in the exams and gain confidence in answering them.
Quick Revision Tool: Our NCERT solutions serve as an excellent tool for last-minute revision. The solutions cover all key points, definitions, and explanations, ensuring that you have everything you need to quickly review before exams.

Importance of Structured Answers for Scoring Higher Marks

In exams, it's not just about getting the right answer—it's also about presenting it in a well-structured and logical way. Our NCERT solutions for Class are designed to guide you on how to write answers that are organized and effective for scoring high marks.

Precise and Concise Answers: Our solutions are crafted to provide answers that are to the point, without unnecessary elaboration. This ensures that you don't waste time during exams and focus on delivering accurate answers that examiners appreciate.
Step-Wise Marks Distribution: We understand that exams often allot marks based on specific steps or points. Our NCERT solutions break down each answer into structured steps to ensure you cover all essential points required for full marks.
Improved Presentation Skills: By following the format of our NCERT solutions, you learn how to present your answers in a systematic and logical manner. This helps in making your answers easy to read and allows the examiner to quickly identify key points, resulting in better scores.
Alignment with NCERT Guidelines: Since exams are often set in alignment with NCERT guidelines, our solutions are tailored to follow the exact format and language that is expected in exams. This can improve your chances of scoring higher by meeting the examiner's expectations.