NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

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Many students are intimidated by the subject of Mathematics. The concepts, formulas, theorems, and calculations of the subject are challenging for them. Students should have a thorough understanding of the topics in Mathematics since it is a purely conceptual subject. Chapter 8 Class 10 Mathematics is Trigonometry. Students may find it difficult to understand the concepts and calculations in this chapter at first. However, the chapter’s concepts eventually become easier for them with meticulous practise and proper guidance. The curriculum of Chapter-8, Class 10 Mathematics includes the following topics – The Introduction of the Chapter, Trigonometric Ratios, Trigonometric Ratios of Some Specific Angles, Trigonometric Ratios of Complementary Angles, Trigonometric Identities and summary. This is one of the most challenging chapters for the students of Class 10, as it is very difficult for them to grasp the concepts of the chapter. To ensure that students perform effectively in their board examinations, Extramarks provides them with all the essential resources, such as NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to make their learning experience comprehensive and effortless.

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NCERT Textbooks do not include the solutions to the questions mentioned in the respective exercises of the listed chapters. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 provided by Extramarks, can assist students in understanding the questions provided in the NCERT textbooks while providing a clear approach and solution to all the questions. The students’ basic mathematical concepts arestructured with the assistance of NCERT. Therefore, students should thoroughly practise the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. In addition, students who have problems with calculation or have difficulty understanding the concepts in the chapter will greatly benefitfrom practising these solutions. Therefore, Extramarks provides learners with the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1, so as to facilitate their conceptual understanding and help them perform well in their Class 10 Mathematics board examinations.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Ex 8.1) Exercise 8.1

The Central Board of Secondary Education is a major board of education under the control of the Government of India. In India, the Central Board of Secondary Education controls, manages, and oversees the school education system and matters related to the academic curriculum. All the schools that are associated with CBSE, especially Classes 9 to 12, follow the NCERT curriculum. Each year, between February and May, CBSE conducts board examinations for Classes 10 and 12. The NCERT curriculum is followed by all schools affiliated with the CBSE board. In 1961, the Indian Government established NCERT as an independent organisation to assist the Central and State Governments in improving the quality of education in schools. In the areas of education, NCERT is primarily responsible for controlling and coordinating research education, publishing NCERT textbooks, and multimedia digital materials, preparing supplementary material, and much more to help students achieve a successful academic career.Teachers are also trained in innovative educational practises through the organisation’s training programs. Additionally, NCERT conducts bilateral cultural exchanges with other countries in the field of school education, as well as research, development, training, extension, publication, and dissemination. The NCERT also collaborates and interacts with international organisations, visits foreign delegations, and provides various training facilities for educators in developing countries.

CBSE schools follow a completely NCERT-based curriculum, so students should have a solid understanding of the basic concepts over the course of the academic year. Educators who are highly experienced with exceptional expertise in their respective fields prepare the NCERT textbooks. In order to build the basic concepts of the students, these textbooks are the best mode of education. However, since these textbooks do not contain the answers to the questions incorporated in them, Extramarks provides students with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Students may have difficulty finding accurate answers to the questions incorporated in the NCERT textbooks. In order to provide students with the best quality education, Extramarks provides authentic NCERT Solutions for all subjects and classes. Extramarks has taken this initiative to keep the community of learners flourishing so that they can have bright academic careers.

Class 10 Maths Chapter 8 Exercise 8.1: Topics Covered

The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 encompass basic concepts of the specific functions of angles and their application to calculations. It is one of the most used concepts in the fields of Astronomy and Geography. The concept is further applicable in the fields of, Satellite Navigation, Chemistry Number Theory, Electronics, Medical Imaging, Electrical Engineering, Architecture, Mechanical Engineering, Civil Engineering, Oceanography, Seismology, and so on and so forth. Students can refer to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to have a deeper understanding of the topics of the chapter. Chapter 8 Class 10 Mathematics includes the following topics- Introduction of the Chapter, and Trigonometric Ratios. Even though there are a limited number of topics in the exercise, students may find it difficult to comprehend the concepts. To assist students with the preparation for board and competitive examinations, NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 are an essential resource. Furthermore, students who want to pursue Mathematics for further studies or want to perform well in any competitive examination should review the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 very keenly.

It has always been the objective of Extramarks to promote academic excellence among students. It provides students with all the essential learning resources that are required to obtain outstanding results in any in-school, board, or competitive examination. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 provided by Extramarks are compiled by the in-house mathematics mentors of Extramarks, who are well-qualified and greatly experienced. These solutions are curated in an easy-to-understand language and are also very well structured so that learners can easily understand them. Students who want to improve their mathematical skills will find the NCERT Solutions for Class 10 Maths, Chapter 8, Exercise 8.1 to be extremely beneficial.These solutions are a reliable resource for Class 10 mathematics board exam preparation.What is Trigonometry?

Trigonometry is the study of the relationship between the sides and angles of triangles, as well as the relevant functions of angles. Students can refer to NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 for a better understanding of Chapter 8 Trigonometry. In order to prepare students effectively for board examinations, Extramarks offers them a variety of learning modules. Extramarks is a paradigm of the benefits of e-learning. K12 Study Material provided by Extramarks helps students score well in examinations by providing complete and convenient study materials to them. Through Extramarks, students are able to self-assess their performance and track their academic progress through In-depth Performance Reports. In Live Doubt-Solving classes, mentors of Extramarks provide students with the facility of attending interactive sessions. As a result, they can perform well in the examinations by resolving their queries right away. To clear up their doubts, students can also refer to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Extramarks gives students access to chapter-based worksheets, unlimited practise questions, interactive activities, and more, so they can become proficient in all subjects. A Visual Learning Journey is also provided by the Learning App so that students can quickly grasp and engage with their subject’s curriculum. Educators at Extramarks provide students with the best possible guidance in their respective fields.

What are Trigonometric Identities?

Trigonometric Identities are a set of equalities involving trigonometry functions that hold true for all values of variables arranged in an equation. Each trigonometric identity is based on trigonometric ratios. The trigonometric ratios are cosine, tangent, sine, secant, cosecant, and cotangent. The sides of the right triangle are used to define these trigonometric ratios. Students can use NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to develop the foundational skills they need to solve questions accurately and efficiently in their examinations. A comprehensive course is provided by Extramarks to learners to ensure their success in examinations. The Extramarks Learning Appis available for convenient download. This application is easy to use and a great online learning tool. Students can refer to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 provided by Extramarks for a better understanding of the chapter.

What Are Trigonometry Ratios?

In a right-angled triangle, trigonometric ratios represent the values of all trigonometric functions. The trigonometric ratios of a right-angled triangle are the ratios of its sides to its acute angles. The adjacent side is the base of the right triangle, and the opposite side is perpendicular. With the help of the Extramarks website, students can easily access NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 at any time and from any place. A simple language has been used to compile these solutions. A comprehensive understanding of the basic concepts of a curriculum is essential for students to do well in a subject. In order to make learning easier for students, Extramarks provides them with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. In order to master the basics of various mathematical themes, the student should familiarise themselves with the NCERT textbook. Comprehensive step-by-step explanations and detailed answers are provided on the Extramarks website to help students better understand the concepts.Students can rely on these solutions since they have been compiled by subject matter experts of Extramarks. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 are provided on Extramarks, a website that provides reliable tools for preparing for board examinations.

For teaching students the concepts of Class 10 Maths Chapter 8 Exercise 8.1, schools could use these NCERT Solutions as efficient digital teaching aids. These solutions could be used by students in preparation for examinations. Schools heavily rely on the NCERT curriculum for instruction since board examinations are centred around an evaluation of its fundamentals. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 should therefore be thoroughly reviewed by students. In order to have a better understanding of the concepts and questions in Class 10 Mathematics Chapter 8, students should practise the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. In order to gain an in-depth understanding of concepts, it is crucial to practice the questions in NCERT textbooks thoroughly. For students to achieve academic success, Extramarks offers a variety of learning tools. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 are one such learning resource. Prior to their board examinations, students should practise these solutions in order to succeed.

Access NCERT Solutions for Class -10 Maths Chapter 8 – Introduction to Trigonometry

To make learning easy and systematic, students need comprehensive study materials. Extramarks provides students with the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 that are properly organised answers to the questions provided in the NCERT textbooks. These solutions assist students in understanding the chapter better. Furthermore, students will save time by not having to search elsewhere for NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. These are step-by-step solutions that are well explained and very easy to access. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 and other resources that help students achieve high scores in board examinations are available on Extramarks. The practise of these solutions increases the academic efficiency of students by helping them gain a better understanding of the concepts of the chapter. Students can find the most straightforward NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 on the Extramarks website.

It can be challenging for students to learn new and complicated concepts. There are many new concepts that cause them to have doubts and inquiries. As Mathematics is primarily conceptual, understanding all the concepts in each chapter is crucial. Each chapter contains so many exercises that introduce new properties, operations, and concepts. Students could have a hard time remembering all these different concepts. For this reason, Extramarks provides students with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to facilitate their understanding of the chapter. In addition, Extramarks offers learning modules such as K12 study Materials, Live Doubt-Solving Sessions, Assessment Centres, etc. Keeping track of topics covered allows students to keep up with topics they have already covered and those they need to practise more. This enables students to achieve their goals and obtain good results in their board examinations. Since NCERT Textbooks do not contain solutions, students have difficulty finding accurate and detailed answers to the questions. Searching for answers on various portals or going through extensive notes can be time-consuming. On Extramarks, students can find solutions to a variety of other chapters, including NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

Practising NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 will help students learn the basics of the chapter and improve their calculation speed. These solutions focus on a specific sphere of the chapter. A variety of learning tools are provided by Extramarks to students, including K12 Study materials for board examinations, complete syllabus coverage, media-rich content, curriculum-mapped learning, and much more, so they can learn systematically. Additionally, the website provides students with NCERT Solutions for all academic disciplines and classes, along with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Students can refer to the Extramarks learning portal to access the NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2, NCERT Solutions Class 1 and a variety of other reliable study material. The e-learning platform provides students with the opportunity to learn, practice, and clarify their doubts. In order to ensure a smooth learning experience, students should comply with the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 designed by knowledgeable mentors of Extramarks. To understand the concepts of the chapter, learners should review Extramarks’ NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1.

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Exercise 8.1 Class 10 Solutions Chapter 8 Trigonometry- Free PDF Download

Practising the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 is a very essential exercise for the quick revision of crucial concepts prior to the board examinations. Students benefit from these solutions, as they help them to score high in Mathematics. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 also assist students in gaining a deeper understanding of the topic. In addition, students will be able to maintain a better pace for their Mathematics board examinations if they practise these questions. These solutions assist students in solving lengthy questions more accurately. Before appearing in their board examinations, Extramarks recommends students review the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. In subjects like Mathematics, revision is a very important part of preparation. In order for students to succeed in their examinations, Extramarks provides them with the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Students can create a revision schedule based on the chapter notes they prepare by using these solutions. Additionally, the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 are detailed solutions to help students succeed in their examinations.

Class 10th Maths Chapter 8 Exercise 8.1

The Class 10 board examinations are of great importance to students’ careers, and the first step they must take is to review the Mathematics syllabus. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 provided by Extramarks is designed to assist students in learning the concepts covered in Chapter 8. As the subjects of Class 10 form the foundation for higher education, students must have a deep understanding of the topics covered in the course of their curriculum. The tenth grade marks an important milestone in the academic career of a student. Duringthis academic year, they formulate their career mindsets. The Class 10 board examinations are therefore very important for them. In addition to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1, Extramarks provides students with several sample question papers, revision notes, important questions, solutions, and much more. As a result, students will be able to practise thoroughly and score highly in the board examinations.

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NCERT Solutions for Class 10 Maths Chapter 8 All Other Exercises

Students can improve their knowledge of Chapter 8-Trigonometry by practising NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Apart from these solutions, the other exercises of the chapter help students to understand the fundamentals of trigonometry and also assist them in applying those concepts practically. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 and the other exercises are designed for students to improve their mathematical concepts and calculation abilities efficiently. By going through these exercises, students can practise the concepts and functions of the chapter, which are very essential to succeed in any in-school, board, or competitive examinations. Practising the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 and the other exercises is the most important step that students should take in preparation for the chapter their examinations.

Q.1

\begin{array}{l} In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. \\ Determine : \\ (i) sin A, cos A \\ (ii) sin C, cos C \end{array}

Ans

\begin{array}{l} Using Pythagoras Theorem in Δ ABC, we get \\ AC = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 cm \end{array}

\begin{array}{l} (i) \\ sin A = \frac{side opposite to angle A}{hypotenuse} = \frac{BC}{AC} = \frac{7}{25} \\ \cos A = \frac{side adjacent to angle A}{hypotenuse} = \frac{AB}{AC} = \frac{24}{25} \\ (ii) \\ sin C = \frac{side opposite to angle C}{hypotenuse} = \frac{AB}{AC} = \frac{24}{25} \\ \cos C = \frac{side adjacent to angle C}{hypotenuse} = \frac{BC}{AC} = \frac{7}{25} \end{array}

Q.2

In the following figure, tanP - cotR .

Ans

\begin{array}{l} Using Pythagoras theorem in Δ PQR, we get \\ QR = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = \sqrt{25} = 5 cm \end{array}

\begin{array}{l} tanP = \frac{side opposite to angle P}{side adjacent to angle P} = \frac{QR}{PQ} = \frac{5}{12} \\ cotR = \frac{side adjacent to angle R}{side opposite to angle R} = \frac{QR}{PQ} = \frac{5}{12} \\ tanP - cotR = \frac{5}{12} - \frac{5}{12} = 0 \end{array}

Q.3

If sin A = \frac{3}{4}, calculate \cos A and \tan A.

Ans

Let Δ ABC is a right-angled triangle right angled at B.

\begin{array}{l} Given that, \\ sin A = \frac{3}{4} \\ or \frac{BC}{AC} = \frac{3}{4} \\ Let BC be 3k. Therefore, AC will be 4k, where k is \\ a positive integer. \\ Applying Pythagoras Theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {or (4k)}^{2} = {AB}^{2} + {(3 k)}^{2} \\ or 16 k^{2} - 9 k^{2} = {AB}^{2} \\ or 7 k^{2} = {AB}^{2} \\ or   AB = \sqrt{7} k \\ \cos A = \frac{side adjacent to angle A}{hypotenuse} = \frac{AB}{AC} = \frac{\sqrt{7} k}{4 k} = \frac{\sqrt{7}}{4} \\ \tan A = \frac{side opposite to angle A}{side adjacent to angle A} = \frac{BC}{AB} = \frac{3 k}{\sqrt{7} k} = \frac{3}{\sqrt{7}} \end{array}

Q.4

Given 15 cot A = 8, find sin A and sec A.

Ans

Let Δ ABC is a right-angled triangle right angled at B.

\begin{array}{l} Given that, \\ 15 cot A = 8 \\ or cot A = \frac{8}{15} \\ or \frac{AB}{BC} = \frac{8}{15} \\ Let AB be 8k. Therefore, BC will be 15k, where k is \\ a positive integer. \\ Applying Pythagoras theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {or AC}^{2} = {(15k)}^{2} + {(8 k)}^{2} \\ {or AC}^{2} = 225 k^{2} + 64 k^{2} \\ {or  AC}^{2} = 289 k^{2} \\ or  AC = 17 k \\ \sin A = \frac{side opposite to angle A}{hypotenuse} = \frac{BC}{AC} = \frac{15 k}{17 k} = \frac{15}{17} \\ \sec A = \frac{hypotenuse}{side adjacent to angle A} = \frac{AC}{AB} = \frac{17 k}{8 k} = \frac{17}{8} \end{array}

Q.5

Given sec θ = \frac{13}{12}, calculate all other trigonometric ratios.

Ans

Let Δ ABC is a right triangle, right-angled at B.

\begin{array}{l} Given that, \\ sec θ = \frac{13}{12} \\ or \frac{AC}{AB} = \frac{13}{12} \\ Let AC be 13k. Therefore, AB is 12k, where k is \\ a positive integer. \\ Applying Pythagoras Theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {BC}^{2} = {AC}^{2} - {AB}^{2} \\ {or BC}^{2} = {(13k)}^{2} - {(12 k)}^{2} \\ {or BC}^{2} = 169 k^{2} - 144 k^{2} \\ {or  BC}^{2} = 25 k^{2} \\ or  BC = 5 k \\ sinθ = \frac{side opposite to angle θ}{hypotenuse} = \frac{BC}{AC} = \frac{5 k}{13 k} = \frac{5}{13} \\ cosθ = \frac{side adjacent to angle θ}{hypotenuse} = \frac{AB}{AC} = \frac{12 k}{13 k} = \frac{12}{13} \\ tanθ = \frac{side opposite to angle θ}{side adjacent to angle θ} = \frac{BC}{AB} = \frac{5 k}{12 k} = \frac{5}{12} \\ cotθ = \frac{side adjacent to angle θ}{side opposite to angle θ} = \frac{AB}{BC} = \frac{12 k}{5 k} = \frac{12}{5} \\ cosec θ = \frac{hypotenuse}{side opposite to angle θ} = \frac{AC}{BC} = \frac{13 k}{5 k} = \frac{13}{5} \end{array}

Q.6

\begin{array}{l} If ∠ A and ∠ B are acute angles such that cos A = cos B, \\ then show that ∠ A = ∠ B. \end{array}

Ans

Let us consider a Δ ABC in which CD ⊥ AB.

\begin{array}{l} Given that, \\ cos A = cos B \\ or \frac{AD}{AC} = \frac{BD}{BC} \\ or \frac{AD}{BD} = \frac{AC}{BC} \\ Let \frac{AD}{BD} = \frac{AC}{BC} = k \\ or AD = k BD . .. (1) \\ and \\ AC = k BC . .. (2) \\ Applying Pythagoras Theorem in Δ CAD and Δ CBD, we get \\ {CD}^{2} = {AC}^{2} - {AD}^{2} . .. (3) \\ and \\ {CD}^{2} = {BC}^{2} - {BD}^{2} . .. (4) \\ From equations (3) and (4), we get \\ {AC}^{2} - {AD}^{2} = {BC}^{2} - {BD}^{2} \\ (k {BC)}^{2} - {(k BD)}^{2} = {BC}^{2} - {BD}^{2} \\ or k^{2} {(BC}^{2} - {BD}^{2}) = {BC}^{2} - {BD}^{2} \\ or k^{2} = 1 \\ or k = 1 \\ On putting this value of k in equation (2), we get \\ AC = BC \\ i.e., ∠ A = ∠ B \end{array}

Q.7

If cot θ = \frac{7}{8}, evaluate : (i) \frac{(1 + sinθ) (1 - sinθ)}{(1 + cosθ) (1 - cosθ)}, {(ii) cot}^{2} θ

Ans

Let Δ ABC is a right triangle, right-angled at B.

\begin{array}{l} Given that, \\ cot θ = \frac{7}{8} \\ or \frac{AB}{BC} = \frac{7}{8} \\ Let BC be 8k. Therefore, AB is 7k, where k is \\ a positive integer. \\ Applying Pythagoras Theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {or AC}^{2} = {(7k)}^{2} + {(8 k)}^{2} \\ {or AC}^{2} = 49 k^{2} + 64 k^{2} \\ {or  AC}^{2} = 113 k^{2} \\ or  AC = \sqrt{113} k \\ sinθ = \frac{side opposite to angle θ}{hypotenuse} = \frac{BC}{AC} = \frac{8 k}{\sqrt{113} k} = \frac{8}{\sqrt{113}} \\ cosθ = \frac{side adjacent to angle θ}{hypotenuse} = \frac{AB}{AC} = \frac{7 k}{\sqrt{113} k} = \frac{7}{\sqrt{113}} \\ (i) \frac{(1 + sinθ) (1 - sinθ)}{(1 + cosθ) (1 - cosθ)} = \frac{1 - \sin^{2} θ}{1 - \cos^{2} θ} = \frac{1 - {(\frac{8}{\sqrt{113}})}^{2}}{1 - {(\frac{7}{\sqrt{113}})}^{2}} = \frac{113 - 64}{113 - 49} = \frac{49}{64} \\ {(ii) cot}^{2} θ = {(\frac{7}{8})}^{2} = \frac{49}{64} \end{array}

Q.8

If 3 cot A = 4, check whether \frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \cos^{2} A - \sin^{2} A or not.

Ans

Let Δ ABC is a right triangle, right-angled at B.

\begin{array}{l} Given that, \\ cot A = \frac{4}{3} \\ or \frac{AB}{BC} = \frac{4}{3} \\ Let BC be 3k. Therefore, AB = 4k, where k is \\ a positive integer. \\ Applying Pythagoras theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {or AC}^{2} = {(4k)}^{2} + {(3 k)}^{2} \\ {or AC}^{2} = 16 k^{2} + 9 k^{2} \\ {or  AC}^{2} = 25 k^{2} \\ or  AC = 5 k \\ \sin A = \frac{side opposite to angle A}{hypotenuse} = \frac{BC}{AC} = \frac{3 k}{5 k} = \frac{3}{5} \\ \cos A = \frac{side adjacent to angle A}{hypotenuse} = \frac{AB}{AC} = \frac{4 k}{5 k} = \frac{4}{5} \\ \tan A = \frac{side opposite to angle A}{side adjacent to angle A} = \frac{BC}{AB} = \frac{3 k}{4 k} = \frac{3}{4} \\ Now, \\ \frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \frac{1 - {(\frac{3}{4})}^{2}}{1 + {(\frac{3}{4})}^{2}} = \frac{7}{25} \\ \cos^{2} A - \sin^{2} A = {(\frac{4}{5})}^{2} - {(\frac{3}{5})}^{2} = \frac{7}{25} \\ ∴ \frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \cos^{2} A - \sin^{2} A \end{array}

Q.9

\begin{array}{l} In triangle ABC, right-angled at B, if tan A = \frac{1}{\sqrt{3}}, find the value of: \\ (i) sin A cos C + cos A sin C \\ (ii) cos A cos C - sin A sin C \end{array}

Ans

\begin{array}{l} Given that, \\ tan A = \frac{1}{\sqrt{3}} \\ or \frac{BC}{AB} = \frac{1}{\sqrt{3}} \\ Let BC be k . Then, AB = \sqrt{3} k, where k is \\ a positive integer. \\ Applying Pythagoras Theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {or AC}^{2} = {(\sqrt{3} k)}^{2} + {(k)}^{2} \\ {or AC}^{2} = 3 k^{2} + k^{2} \\ {or  AC}^{2} = 4 k^{2} \\ or  AC = 2 k \\ \sin A = \frac{side opposite to angle A}{hypotenuse} = \frac{BC}{AC} = \frac{k}{2 k} = \frac{1}{2} \\ \cos A = \frac{side adjacent to angle A}{hypotenuse} = \frac{AB}{AC} = \frac{\sqrt{3} k}{2 k} = \frac{\sqrt{3}}{2} \\ \sin C = \frac{side opposite to angle C}{hypotenuse} = \frac{AB}{AC} = \frac{\sqrt{3} k}{2 k} = \frac{\sqrt{3}}{2} \\ \cos C = \frac{side adjacent to angle C}{hypotenuse} = \frac{BC}{AC} = \frac{k}{2 k} = \frac{1}{2} \\ Now, \\ (i) \sin A \cos C + \cos A \sin C = \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = 1 \\ (ii) \cos A \cos C - \sin A \sin C = \frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2} = 0 \end{array}

Q.10

\begin{array}{l} In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. \\ Determine the values of sin P, cos P and tan  P. \end{array}

Ans

\begin{array}{l} Given that, \\ PR + QR = 25 cm \\ and \\ PQ = 5 cm \\ Let PR be x . Then, QR = 25 - x \\ Applying Pythagoras Theorem in Δ PQR, we get \\ {PR}^{2} = {PQ}^{2} + {QR}^{2} \\ or x^{2} = {(5)}^{2} + {(25 - x)}^{2} \\ or x^{2} = 25 + 625 + x^{2} - 50 x \\ or 50 x = 650 \\ or x = 13 \\ ∴ PR = x = 13 cm \\ and \\ QR = 25 - x = 25 - 13 = 12 cm \\ \sin P = \frac{side opposite to angle P}{hypotenuse} = \frac{QR}{PR} = \frac{12}{13} \\ \cos P = \frac{side adjacent to angle P}{hypotenuse} = \frac{PQ}{PR} = \frac{5}{13} \\ \tan P = \frac{side opposite to angle P}{side adjacent to angle P} = \frac{QR}{PQ} = \frac{12}{5} \end{array}

Q.11

\begin{array}{l} State whether the following are true or false. Justify \\ your answer. \\ (i) The value of tan A is always less than 1. \\ (ii) sec A = \frac{12}{5} for some value of angle A. \\ (iii) cos A is the abbreviation used for the cosecant \\ of angle A. \\ (iv) cot A is the product of cot and A. \\ (v) sin θ = \frac{4}{3} for some angle θ . \end{array}

Ans
(i) Let us consider the following Δ ABC, right-angled at B.

\begin{array}{l} \tan A = \frac{side opposite to angle A}{side adjacent to angle A} = \frac{12}{5} > 1 \\ ∴ \tan A > 1 \\ So, tan A<1 is not always true. \\ Hence, the given statement is false. \end{array}

(ii) Let us consider the following Δ ABC, right-angled at B.

\begin{array}{l} We consider on the above Δ ABC, right-angled at B. \\ \sec A = \frac{hypotenuse}{side adjacent to angle A} = \frac{AC}{AB} = \frac{12}{5} \\ Now, by using Pythagoras theorem, we have \\ {BC}^{2} = {AC}^{2} - {AB}^{2} = 144 - 25 = 119 \\ or BC = \sqrt{119} = 10 .9 \\ Therefore, \sec A = \frac{12}{5} is possible for some values \\ of angle A. \\ Hence, the given statement is true. \\ (iii) \\ The abbreviation used for the cosecant of angle A is cosec A. \\ Therefore, the given statement is false. \\ (iv) \\ cot A is not the product of cot and A. Therefore, the given \\ statement is true. \\ (v) \\ \sin θ = \frac{side opposite to angle θ}{hypotenuse} \\ In a right-angled triangle hypotenuse is the longest side. \\ Thus, \\ \sin θ = \frac{side opposite to angle θ}{hypotenuse} < 1 \\ Therefore, \sin θ = \frac{4}{3} > 1 is not possible for any value of θ . \\ Hence, the given statement is false. \end{array}

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FAQs (Frequently Asked Questions)

1. Where can students find the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1?

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The first and foremost step that students must take for the preparation of their Class 10 Mathematics board examination is to thoroughly go through the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Thereafter, they should review the important questions, extra questions and revision notes of the chapter’s curriculum. As a result, students would be rendered able to determine their preparation level as well as their strong and weak areas with regard to the curriculum. As these tools highlight the key points of the chapter, they are extremely useful for students when they need to revise quickly. In addition, they should thoroughly review past years’ papers and sample papers of the subject’s curriculum. It is through this process that students will develop a strong conceptual understanding of the subject and be able to solve any complicated problems they may encounter during their board examinations. Students can also benefit from sample papers and past year’s papers in order to get a better understanding of the examination pattern, which will help them to avoid last-minute challenges and perform better in their examinations. Students are advised to be thorough with the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 before attempting their Mathematics board examinations.

4. Would it be necessary to practice all the questions in NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1?

Yes, it is crucial for the learners to review all the NCERT questions of the Class 10 Maths Ch 8 Ex 8.1. Each question contains a unique fact or concept related to the chapter. Moreover, the question paper of the Mathematics Class 10 board examination is also based on the NCERT curriculum. Thus, any NCERT question from the chapter may appear in the board examinations. In addition, NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 assist students in strengthening their fundamentals so that they are better prepared to solve any complicated questions that may appear in the board examinations.

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Click on the given link to download the Class 10 Maths Chapter 8 Exercise 8.1 Solutions.

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Ex 8.1) Exercise 8.1