# NCERT Solutions For Class 10 Maths Chapter 8 Introduction To Trigonometry (Ex 8.2) Exercise 8.2

Both public and private schools in New Delhi receive secondary education from the Central Board of Secondary Education. The Indian government oversees and manages the CBSE, a national education system. To raise the standard of schooling in India, the National Council of Educational Research and Training was established in 1961. These groups provide guidance and support to the state and federal governments. Additionally, they create and disseminate educational goods like kits, textbooks, additional materials, newsletters, journals, and multimedia items.

Class 10 is a crucial stage in a student’s academic career. The grades they earn on the board exams will determine their destiny. Depending on their grades and interests, students can pick between Humanities, Commerce, and Science. Without a question, Mathematics is a challenging subject, but it is also rewarding. When practised correctly and with complete focus, mathematical ideas are not difficult to master. Students must receive at least a 33% on their theory and practical exams in order to advance to Class 11 and Class 12.

Mathematics explains common patterns that students come across in daily life, where successive terms are obtained by multiplying the previous terms by a predetermined quantity. Additionally, they will learn how to use this information to solve some common issues.

In India, a major organisation called the National Council for Educational Research and Training, or NCERT, performs its duties. The Indian Government established NCERT in 1961 because of the requirement for a group of highly trained individuals who had been hired specifically to study and identify the best answers to all problems relating to the nation’s educational progress. On issues pertaining to the educational development of the nation’s youth, NCERT provided support to both the Central Government and the State Governments.

NCERT and CBSE have a very synergistic connection in which they both rely heavily on one another. Students must adhere completely to the norms and guidelines that NCERT publishes on a regular basis. Additionally, NCERT has its own publishing firm through which it releases books based on the carefully selected curriculum and the approved syllabus. Teachers universally instruct all of their students to follow the NCERT textbook because it directly follows their norms and recommendations. Since these books are totally in line with the NCERT syllabus, students should treat the NCERT textbook as their main text because practising problems and questions from these volumes would adequately prepare them for the tests.

If students want a quick review of the various tips and tricks explained in the NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2, they can download the PDF format from the Extramarks website. The NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 are available in PDF format and can be downloaded or printed according to the student’s convenience. This chapter’s goal is to provide students with an understanding of Trigonometry. Students can save the PDF file of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 on their devices so that they do not have to worry about internet connectivity while preparing for the test. Students can download the solutions for Class 10 Maths Ch 8 Ex 8.2 in PDF format for offline access via the Extramarks website.

**NCERT Solutions For Class 10 Maths Chapter 8 Introduction To Trigonometry (Ex 8.2) Exercise 8.2**

Trigonometry is a branch of Mathematics that plays an important role in many professions. Simply, Trigonometry is the study of triangles and their side lengths and angles.

As one of the most important areas of Mathematics, especially for careers based on calculating angles, working knowledge of Trigonometry and how to use it is important for students of all ages. The solutions to Maths Class 10 Chapter 8 Exercise 8.2 provide the students with a thorough understanding of Trigonometry and clarify their doubts to boost their preparation for final exams.

Extramarks online tutorials of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 make learning Trigonometry easy for students of all ages. Through consistent learning, students master all aspects of Trigonometry and improve their Mathematics skills. Trigonometry is the study of triangles, more specifically the angles and dimensions of triangles. Trigonometry is an integral part of modern engineering, design, architecture, and other fields.

Using Trigonometry skills, students can calculate the exact angles of the sides of a triangle, the distances between different points of the triangle, and other information that is important in a variety of situations. Trigonometry skills play an important role in many different professions, such as architecture and engineering. For this reason, understanding Trigonometry is important for students interested in science or engineering.

Learning Trigonometry involves learning how to calculate the angles and dimensions of specific shapes using trigonometric functions such as the sine and cosine of angles. An effective Trigonometry course should focus on teaching students the most important trigonometric functions. One should also use these features in hands-on exercises to help students develop their skills. Students must focus on Mathematics Class 10 Chapter 8 exercise 8.2. Extramarks lets students learn important trigonometric functions through a series of lessons of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 updated on the website of Extramarks.

Trigonometry is a complex subject that takes a long time for students to master because the functions involved in Trigonometry are very different from problem-solving techniques used in other subjects and many students need the practice to progress this topic. Learning Trigonometry allows students to improve their Mathematics skills in an environment that encourages focus, practise, and repetition of revision. This makes it easier for students to overcome difficulties and acquire a comprehensive knowledge of Trigonometry. Having a solid knowledge of Geometry and Algebra is essential as these form the foundation for learning Trigonometry. Trigonometry, like any other branch of Mathematics, requires practise in the various trigonometric functions. Mastering Trigonometry requires understanding how the various trigonometric functions work and how they can be used. This takes focus and practice. Therefore, it is important for students to participate in Trigonometry research. Extramarks tutorials on NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 can be used alongside the course as a supplement or as part of a classroom Trigonometry lesson.

As one of the most important areas of Mathematics, Trigonometry should be the focus of every student. Good Trigonometry skills enable students to calculate complex angles and dimensions in a relatively short amount of time.

NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 is useful for students as it helps them understand the concepts and do well on the Class 10 CBSE exam. NCERT solutions are designed and reviewed by subject matter experts covering all textbook issues. These NCERT solutions are designed following the latest updates to the CBSE syllabus for 2022-23 and guidelines based on exam patterns.

The NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 provide a strong foundation for each concept found in all chapters. Students will be able to clear their doubts and understand the basics contained in this chapter. Students can also solve difficult problems in each exercise with the help of these NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2.

It is important to practise all the problems in NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2. Each answer provided is an application of a different concept, and practising these questions will help students understand all the concepts. Students can refer to NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 or they can download it from the Extramarks website or mobile application. Extramarks offers NCERT Solutions for different classes, such as NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, and NCERT Solutions Class 6. The time it takes to complete solving questions with the help of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 will vary for each student. It may take several hours or a day or two to complete the entire exercise. Students may be able to complete the exercises in less time if the concepts are very clear. Therefore, to support students, Extramarks provides access to curated answers to these questions from NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2. Students can examine NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 to clear their doubts. All resources and NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 are available on the Extramarks’ website or mobile application.

**Important Points To Remember Before Solving Exercise 8.2**

Students are already aware of the configurations for angles 30 degrees, 45 degrees, 60 degrees and 90 degrees by studying Geometry. In this section, students will find the trigonometric ratio values for these angles and 0 degrees. The trigonometric ratio of 45°, the the triangular ratio of 30° and 60° and the triangular ratio of 0° and 90° are explained with good examples.

- Trigonometric Ratios Of Complementary Angles

Two angles are complementary if the sum of the two angles is 90°. This topic describes various formulas for solving numerical problems related to trigonometric ratios. Students must refer to the NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 for the perfect solutions.

- Trigonometric Identities

Students may recall that an equation is said to have an identity if it holds for all values of the variables involved. Similarly, an equation involving trigonometric ratios of angles is called a trigonometric identity if it holds for all values of the angles involved. In this section, students will prove the trigonometric identity and use it to prove other useful trigonometric identities.

- Summary

This summary is a conceptual summary containing all the key points students need to remember to solve the numerical problems associated with the chapter. Students must expect at least a few mandatory questions in this chapter. The main topics related to the chapter are:

In previous classes, students learned about triangles, specifically right triangles. In this chapter, they will study the ratio of the sides of a right triangle to its Acute Angle (called the triangular angle ratio). It also defines the trigonometric ratio of the 0° and 90° measurement angles. Calculating the trigonometric ratios of some specific angles and establishing some identities involving these ratios. This is called the trigonometric identity.

Students learned the concepts of relationships in previous classes. Now the chapter defines certain ratios that affect the sides of a right triangle and calls them trigonometric ratios. Using suitable examples, the topic is explained using various functions of Trigonometry.

The CBSE Class 10 Mathematics exam consists of various questions that carry a total weightage of 12 marks from Unit 5 ‘Trigonometry’ out of a total of 80 marks. This book has four parts. Each part has a different score, and the questions are 1 mark, 2 marks and 3 marks.

**1. Trigonometric Ratios**

Sine (sin), Cosine (cos), Tangent (tan), Cotangent (cot), Cosecant (cosec), and Secant (sec) are the 6 trigonometric ratios. In Geometry, Trigonometry is the branch of Mathematics that deals with the sides and angles of right triangles. Therefore, trigonometric ratios are evaluated in terms of sides and angles.

Note: The opposite side is the vertical side, and the adjacent side is the base of the right triangle. Also, see Trigonometric Functions for more information on these ratios or functions. triangle identity

** 2. Standard Values Of Trigonometric Ratios [Include a Table] [Include Download PDF Button]**

Sin 30° = ½

Cos 90° = 0

Tan 45° = 1

Angle | 0° | 30° | 45° | 60° | 90° |

Sin C | 0 | 1/2 | 1/√2 | √3/2 | 1 |

Cos C | 1 | √3/2 | 1/√2 | 1/2 | 0 |

Tan C | 0 | 1/√3 | 1 | √3 | ∞ |

Cot C | ∞ | √3 | 1 | 1/√3 | 0 |

Sec C | 1 | 2/√3 | √2 | 2 | ∞ |

Cosec C | ∞ | 2 | √2 | 2/√3 | 1 |

**Access NCERT Solutions For Class -10 Maths Chapter 8 – Introduction To Trigonometry**

Students can download a PDF of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 and all the other chapters and their exercises in one place, compiled by expert teachers following the guidelines of the NCERT textbook. The NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 has solutions that help students review the outline and score more points. Students can also download NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 from the Extramarks app.

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**NCERT Solutions For Class 10 Maths Chapter 8 Introduction To Trigonometry Exercise 8.2**

Choosing the NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 is considered to be the best option for CBSE students when it comes to exam preparation. The chapter has many exercises. Extramarks provides the NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2 in PDF format. Students can download these solutions according to their convenience or study online directly from Extramarks’ website or mobile application.

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**Q.1 **

$\begin{array}{l}\text{Evaluate the following:}\\ \text{(i) sin\hspace{0.17em}60\xb0\hspace{0.17em}cos 30\xb0}+\text{sin 30\xb0\hspace{0.17em}cos 60\xb0}\\ {\text{(ii) 2tan}}^{\text{2}}\text{45\xb0}+{\text{cos}}^{\text{2}}\text{30\xb0}-{\text{sin}}^{\text{2}}\text{60\xb0}\\ \text{(iii)}\frac{\text{cos 45\xb0}}{\text{sec 30\xb0}+\text{cosec 30\xb0}}\\ \text{(iv)}\frac{\text{sin 30\xb0}+\text{tan 45\xb0}-\text{cosec 60\xb0}}{\text{sec 30\xb0}+\text{cos 60\xb0}+\text{cot 45\xb0}}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\frac{{\text{5cos}}^{2}\text{60\xb0}+{\text{4sec}}^{2}\text{30\xb0}-{\text{tan}}^{2}\text{45\xb0}}{{\text{sin}}^{2}\text{30\xb0}+{\text{cos}}^{2}\text{30\xb0}}\end{array}$

**Ans**

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}60\xb0\hspace{0.17em}cos 30\xb0}+\text{sin 30\xb0\hspace{0.17em}cos 60\xb0}=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}=1\\ \text{(ii)}\\ {\text{2tan}}^{\text{2}}\text{45\xb0}+{\text{cos}}^{\text{2}}\text{30\xb0}-{\text{sin}}^{\text{2}}\text{60\xb0}=2\times {\left(1\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}-{\left(\frac{\sqrt{3}}{2}\right)}^{2}=2\\ \text{(iii)}\\ \frac{\text{cos 45\xb0}}{\text{sec 30\xb0}+\text{cosec 30\xb0}}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3)}}\times \frac{(1-\sqrt{3)}}{(1-\sqrt{3)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}-3}{2\sqrt{2}(1-3)}=\frac{3-\sqrt{3}}{4\sqrt{2}}\\ \text{(iv)}\\ \frac{\text{sin 30\xb0}+\text{tan 45\xb0}-\text{cosec 60\xb0}}{\text{sec 30\xb0}+\text{cos 60\xb0}+\text{cot 45\xb0}}=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\\ \text{}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \frac{3\sqrt{3}-4}{3\sqrt{3}-4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{(3\sqrt{3}-4)}^{2}}{{\left(3\sqrt{3}\right)}^{2}-16}=\frac{{\left(3\sqrt{3}\right)}^{2}-24\sqrt{3}+16}{27-16}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{43-24\sqrt{3}}{11}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\\ \frac{{\text{5cos}}^{2}\text{60\xb0}+{\text{4sec}}^{2}\text{30\xb0}-{\text{tan}}^{2}\text{45\xb0}}{{\text{sin}}^{2}\text{30\xb0}+{\text{cos}}^{2}\text{30\xb0}}=\frac{5\times {\left(\frac{1}{2}\right)}^{2}+4\times {\left(\frac{2}{\sqrt{3}}\right)}^{2}-1}{{\left(\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}=\frac{67}{12}\end{array}$

**Q.2 **

$\begin{array}{l}\text{Choose the correct option and justify your choice:}\\ \text{(i)}\frac{\text{2tan30\xb0}}{\text{1}+{\text{tan}}^{2}\text{30\xb0}}=\\ \text{(A) sin 60\xb0 (B) cos 60\xb0 (C) tan 60\xb0 (D) sin 30\xb0}\\ \text{(ii)}\frac{\text{1}-{\text{tan}}^{2}\text{45\xb0}}{\text{1}+{\text{tan}}^{2}\text{45\xb0}}=\\ \text{(A) tan 90\xb0 (B) 1 (C) sin 45\xb0 (D) 0}\\ \text{(iii) sin2A}=\text{2sinA is true when A}=\\ \text{(A) 0\xb0 (B) 30\xb0 (C) 45\xb0 (D) 60\xb0}\\ \text{(iv)}\frac{\text{2tan30\xb0}}{1-{\text{tan}}^{2}\text{30\xb0}}=\\ \text{(A) cos 60\xb0 (B)sin 60\xb0 (C) tan 60\xb0 (D) sin 30}\end{array}$

**Ans**

$\begin{array}{l}\text{(i)}\\ \frac{\text{2tan30\xb0}}{\text{1}+{\text{tan}}^{2}\text{30\xb0}}=\frac{2\times \frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}=\frac{2\times 3}{4\sqrt{3}}=\frac{\sqrt{3}}{2}=\mathrm{sin}60\mathrm{\xb0}\\ \text{Hence, the correct option is (A).}\\ \text{(ii)}\\ \text{}\frac{\text{1}-{\text{tan}}^{2}\text{45\xb0}}{\text{1}+{\text{tan}}^{2}\text{45\xb0}}=\frac{1-1}{1+1}=0\\ \text{Hence, the correct option is (D).}\\ \text{(iii)}\\ \text{sin2A}=\text{2sinA}\\ \text{On putting A}=0\mathrm{\xb0},\text{we get}\\ \text{sin\hspace{0.17em}}0\mathrm{\xb0}=\text{2sin\hspace{0.17em}}0\mathrm{\xb0}\\ \text{or 0}=2\times 0\\ \text{or 0}=0\\ \text{Therefore, the correct option is (A).}\\ \text{Other given options does not satisfy the given condition.}\\ \text{(iv)}\\ \frac{\text{2tan30\xb0}}{1-{\text{tan}}^{2}\text{30\xb0}}=\frac{\text{2}\times \frac{1}{\sqrt{3}}}{1-{\left(\frac{1}{\sqrt{3}}\right)}^{2}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3}=\mathrm{tan}\text{\hspace{0.17em}}60\mathrm{\xb0}\\ \text{Therefore, the correct option is (C).}\end{array}$

**Q.3 **

**Ans**

$\begin{array}{l}\text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan}(\mathrm{A}+\mathrm{B})=\sqrt{\text{3}}\\ \text{or tan (A}+\text{B})=\text{tan\hspace{0.17em}60\xb0}\\ \text{or A}+\text{B}=60\mathrm{\xb0}\text{}...\text{(1)}\\ \text{Again, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan}(\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{\text{3}}}\\ \text{or tan (A}-\text{B})=\text{tan\hspace{0.17em}}3\text{0\xb0}\\ \text{or A}-\text{B}=30\mathrm{\xb0}\text{}...\text{(2)}\\ \text{On adding equations (1) and (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 2A}=90\mathrm{\xb0}\\ \text{or A}=45\mathrm{\xb0}\\ \text{On putting this value of A in equation (1), we get}\\ \text{B}=15\mathrm{\xb0}\end{array}$

**Q.4 **

$\begin{array}{l}\text{State whether the following are true or false.}\\ \text{Justify your answer.}\\ \text{(i) sin}(\mathrm{A}+\mathrm{B})=\text{sin A}+\text{sin B.}\\ \text{(ii) The value of sin}\mathrm{\theta}\text{\hspace{0.33em}increases as\hspace{0.33em}}\mathrm{\theta}\text{\hspace{0.33em}increases.}\\ \text{(iii) The value of cos}\mathrm{\theta}\text{\hspace{0.33em}increases as\hspace{0.33em}}\mathrm{\theta}\text{\hspace{0.33em}increases.}\\ \text{(iv) sin}\mathrm{\theta}=\text{cos}\mathrm{\theta}\text{\hspace{0.33em}for all values of\hspace{0.33em}}\mathrm{\theta}.\\ \text{(v) cot A is not defined for A}=0.\end{array}$

**Ans**

$\begin{array}{l}\text{(i)}\\ \text{Let A}=30\mathrm{\xb0}\text{and B}=60\mathrm{\xb0}.\\ \text{Now,}\\ \mathrm{sin}(\text{A}+\text{B)}=\mathrm{sin}(30\mathrm{\xb0}+60\mathrm{\xb0}\text{)}=\mathrm{sin}90\mathrm{\xb0}=1\\ \text{and}\\ \text{sin A}+\text{sin B}=\text{sin}30\mathrm{\xb0}+\text{sin\hspace{0.17em}\hspace{0.17em}}60\mathrm{\xb0}=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\\ \therefore \mathrm{sin}(\text{A}+\text{B)}\ne \text{sin A}+\text{sin B}\\ \text{Hence, the given statement is false.}\\ \text{(ii)}\\ \text{We know that}\\ \mathrm{sin}0\mathrm{\xb0}=0,\\ \mathrm{sin}30\mathrm{\xb0}=\frac{1}{2}=0.5,\\ \mathrm{sin}45\mathrm{\xb0}=\frac{1}{\sqrt{2}}=0.7,\\ \mathrm{sin}60\mathrm{\xb0}=\frac{\sqrt{3}}{2}=0.87,\\ \mathrm{sin}90\mathrm{\xb0}=1\\ \text{We see that the value of sin}\mathrm{\theta}\text{increases as}\mathrm{\theta}\text{increases}\\ \text{in the interval}0\mathrm{\xb0}\le \mathrm{\theta}\le 90\mathrm{\xb0}.\\ \text{Hence, the given statement is true if}\mathrm{\theta}\text{lies in the}\\ \text{interval}0\mathrm{\xb0}\le \mathrm{\theta}\le 90\mathrm{\xb0}.\\ \text{(iii)}\\ \text{We know that}\\ \mathrm{cos}0\mathrm{\xb0}=1,\\ \mathrm{cos}30\mathrm{\xb0}=\frac{\sqrt{3}}{2}=0.87,\\ \mathrm{cos}45\mathrm{\xb0}=\frac{1}{\sqrt{2}}=0.7,\\ \mathrm{cos}60\mathrm{\xb0}=\frac{1}{2}=0.5,\\ \mathrm{cos}90\mathrm{\xb0}=0\\ \text{We see that the value of}\mathrm{cos\theta}\text{decreases as}\mathrm{\theta}\text{increases}\\ \text{in the interval}0\mathrm{\xb0}\le \mathrm{\theta}\le 90\mathrm{\xb0}.\\ \text{Hence, the given statement is false if}\mathrm{\theta}\text{lies in the}\\ \text{interval}0\mathrm{\xb0}\le \mathrm{\theta}\le 90\mathrm{\xb0}.\\ \text{(iv)}\\ \text{We know that}\\ \mathrm{sin}0\mathrm{\xb0}=0\text{and}\mathrm{cos}0\mathrm{\xb0}=1,\\ \mathrm{sin}30\mathrm{\xb0}=\frac{1}{2}=0.5\text{and}\mathrm{cos}30\mathrm{\xb0}=\frac{\sqrt{3}}{2}=0.87.\\ \text{Therefore, sin}\mathrm{\theta}\ne \text{cos}\mathrm{\theta}\text{for all values of}\mathrm{\theta}.\\ \text{Hence, the given statement is false}.\\ \text{(v)}\\ \text{We know that cot A}=\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}.\text{}\\ \text{So, for A}=0,\text{we have}\\ \text{cot 0}=\frac{\mathrm{cos}\text{0}}{\mathrm{sin}\text{0}}=\frac{1}{0}\\ \text{We know that division by 0 is not defined. Therefore, cot 0}\\ \text{is not defined.}\\ \text{Hence, the given statement is true}.\end{array}$

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