NCERT Solutions for Class 11 Maths Chapter 3- Trigonometric Functions Exercise 3.3

The academic discipline of Mathematics has incomparable significance in the daily lives of humans. It is an applied science that is innately associated with a multitude of lucrative careers as well as higher education opportunities. Undoubtedly, Mathematics is one of the most logically coordinated and significant scientific disciplines. India, in particular, has a long history of mathematical innovation and progress.Outstandingly excellent mathematicians have left their imperishable marks on the entire academic arena with their ingenuity and exceptional talent over the ages. Mathematics education has been delivered in the form of an applied science since the early stages of schooling.Accordingly, students are motivated to make consistent efforts to improve their understanding and comprehension of the subject. This is to say that Mathematics exhibits great potential and prominence as an academic subject.

A comprehensive and scientifically arranged academic curriculum for the discipline of Mathematics for students of Class 11 has been prescribed by NCERT. This academic syllabus comprises sixteen different chapters. These chapters have been categorised into six broad units. This classification and the academic content captured within these chapters are the outcomes of painstaking efforts and intensive and eclectic research. The organisation of these chapters into a logically feasible sequence facilitatesfluidity and continuity in the learning process. Themes associated with Trigonometric Functions are part of the third chapter of the academic syllabus prescribed by the NCERT for the students of Class 11 for Mathematics. Trigonometric Functions are a versatile and diverse theme covered by the prescribed academic curriculum of CBSE for students of Class 11. This makes it a prerequisite that the topic of Trigonometric Functions has to be understood in detail to ace exams. The assessments provided as a part of exercise 3.3 Class 11th Maths are holistic and involvequestions that cover a significant portion of this chapter. It includes a variety of sub-topics that are part of the theme of Trigonometric Functions. These themes contain various formulae which have to be comprehended in order to appropriately and adequately solve the problems at hand. The NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 have been designed by closely observing and studying the contributions made by various resourceful and knowledgeable subject experts by Extramarks. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 would be a great resource for learning Mathematics for students of Class 11.

CBSE NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 – Trigonometry

The Extramarks platform for online learning is providing the NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 to cater to the process of learning Mathematics. The NCERT Solutions for Class 11 Maths, Chapter 3 Exercise 3.3, were created with the prescribed academic syllabus for Mathematics for Class 11 students in mind.The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 have been structured in a format which is exactly like the subject-matter arrangement in the NCERT textbook for Mathematics in Class 11. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 have been designed and adjusted in a simple-to-understand framework. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 have been compiled after a careful analysis of past years’ papers and reliable sample question papers. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 could be conveniently used via the medium of the internet for the convenience of students. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 are a source of student-friendly solutions to the assessments provided as a part of Chapter 3 of the NCERT book of Mathematics. Finally, the NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 would be a great addition to the earnest efforts and sincerity of students.

Access NCERT Solutions For Class 11 Maths Chapter 3 – Trigonometric Functions

The academic curriculum for Mathematics prescribed by the NCERT for students in Class 11 is inclusive of the theme of Trigonometric Functions which has been perceived as a complicated and versatile topic.On the other hand, it is a diverse aggregate of conceptually hefty themes. It could also be considered a doorway into a wider horizon of wide-scale research and professional courses, as well as higher education, for thosewho are looking forward to pursuing Mathematics in the future as a career choice. It is noteworthy that students must exhibit a commitment to practising comprehensive and dynamic assessments with the help of high-quality reference materials like the NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 in order to achieve excellence in the topic of Trigonometric Functions. Constant practise of the NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 results in remarkable progress and enhancement in the memory retention and problem comprehension capabilities of students.

NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 Trigonometry-

Additionally, the necessary problem comprehension skills procured by students would enable them to bestow the necessary understanding. Such skills would be of great assistance when students are required to adequately decode a problem at hand in order to develop an appropriate response to it. Students would also need to be skilled enough to select the appropriate practical approach to solving a problem, along with the necessary formulas that have to be applied in order to do the necessary calculations.The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 have been engineered to function as adeptly consumable learning resources during the period of board examinations by the students of Class 11. Therefore, the NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 have been compiled very selectively through the inclusion of the most competent ways of solving descriptive problems related to Trigonometric Functions.

The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 can be used to address a variety of issues and concerns which hinder the process of learning among students. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 could be put to great use by students for recording personal notes for the purpose of examinations. Topic Covered In Class 11 Maths Chapter 3 Exercise 3.3

The theme of Trigonometric Functions occupies the third chapter of the prescribed NCERT textbook for the students of Mathematics in Class 11. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 include solutions for the exercises given as a part of this chapter. This chapter includes a broad multitude of sub-topics and themes which are indispensably crucial from the point of view of board examinations. Moreover, these are also wide and intriguing areas of study which have become invaluable parts of many areas of research and professional occupations. Therefore, it is vital for students to read these topics in comprehensive detail and retain crucial factual and theoretical information as well as important formulae. The Chapter 3 covers a wide range of topics. These include an introduction to trigonometric functions, positive and negative angles, measuring angles in radians and in degrees, and the conversion of one into the other. The definition of trigonometric functions with the help of the unit circle, the truth of the sin2x+cos2x=1, for all x, signs of trigonometric functions, and the domain and range of trigonometric functions have also been encapsulated in the third chapter. Other themes that are part of the chapter on Trigonometric Functions are graphs of trigonometric functions, identities related to sin 2x, cos2x, tan 2x, sin3x, cos3x and tan3x and general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a. The NCERT Solutions for Class 11 Maths, Chapter 3, Exercise 3.3, cover all of these topics and provide appropriate solutions to help students succeed in their exams.Free PDF Download

The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 designed by Extramarks are easily accessible in PDF format. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 can be downloaded by students for quick access to quality academic reference materials. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 could also be used as teaching aids by parents or guardians who make earnest efforts to tutor their children on their own accord. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 are all-inclusive of the formulae which have been applied in order to reach solutions while doing complex calculations. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 consist of solutions that have been curated to adapt to the scheme of mark distribution as it has been conventionalized in the board examinations. NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 – Trigonometry

Trigonometry is quite a distinct field of study if one perceives it comparatively in relation to the academic curriculum of Mathematics which has been a part of the student’s school schedules in previous classes. This is the primary reason why students are quite apprehensive about the conceptual theme of Trigonometry in Mathematics.

Trigonometry refers to the branch of Mathematics that engages with the relationship between the sides and angles of triangles. The history of Trigonometry as an academic sub-discipline can be traced back to the third century BC, when geometry was being applied to astronomical studies. The knowledge based on Trigonometry is practically applied in many professional and academic spheres such as Architecture, Survey, Astronomy, Physics, Engineering and even Crime Scene Investigations. In this path-breaking work, Mathematical Thought From Ancient to Modern Times, Morris Kline has pointed out that while Trigonometry was initially developed in association with astronomy, it was also put to use in the fields of navigation and the crafting of calendars. The foundations of Trigonometry lie in Geometry. Trigonometry is a vital academic concept within Mathematics. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 covers the underlying themes of Trigonometry in detail.

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NCERT Solutions for Class 11 Maths Chapters [Include Chapter wise Pages] also

Mathematics comprises a hugely diverse variety of academic content that is inclusive of elements like formulae, theorems, concepts, principles, etc. These elements are innately associated with each other in a myriad of intriguing ways. However, it is not deniable that these constitutive themes and topics are also distinct in their own right. The nature of the academic component of the scientific discipline of Mathematics necessitates that learners ought to deal with individual problems while simultaneously being aware of the holistically significant concepts on which they are based. Long-drawn calculations, formulae comprising a variety of numerical and other symbols and their concerting derivatives are all characteristic compositions of the applied science of Mathematics. Conventionally established and applied postulates coupled with theorems and working principles could be chronologically dated back thousands of years ago, to the earliest human civilisations in Sumeria and Indus Valley. This timelessness of the discipline has also led it to be a major object of historical interest, research, and analysis. As an expected outcome of the growing interdisciplinary interest, mathematics has acquired the appearance of a sophisticated, complex, and foundational academic discipline. It is widely held to be one of the most fundamentally important scientific disciplines in human society.The application of various mathematical principles allows for the uninterrupted operation of a wide range of human activities.Trigonometry is an inseparable part of the discipline of Mathematics. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 covers the themes and concepts that are part of the Class 11 Maths 3.3 Exercise in detail. Class 11 Maths Chapter 3 Exercise 3.3- Weightage Marks

The prescribed academic curriculum of NCERT for Class 11 for the academic discipline of Mathematics consists of six units. The scheme of marks distribution for these units is as follows- ten marks allotted to Algebra, thirty-five marks for Calculus, fourteen marks for Vectors and 3D Geometry, five marks to Linear Programming and eight marks are being assigned to the last chapter titled Probability. The total counts to eighty marks. Twenty marks are allocated to the practical examination. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 have been designed keeping this in mind. The NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3 would greatly complement the sincere hard work and earnest efforts of students.

Benefits of Maths Chapter 3 Class 11 NCERT Solutions

The process of instruction within a classroom is a remarkably lengthy and complicated one with many specificities that need to be paid more heed to in order to ensure convenient learning. Many causes could be pointed out for this state of affairs. The distinct individual personalities of students, which are characteristic of them as different human beings, could be considered the primary cause. By virtue of this factit becomes evident that students have their own ways of comprehending the subject matter rendered to them. They also choose their own media for conveying their concerns, thoughts and feelings. However, the traditional setting of the classroom uproots all these distinct individuals from their contexts in order to group them together within the classroom. This setting is extremely convenient for imparting education to feasibly large groups of students who are at the same level in terms of intellectual capabilities and comprehension abilities. Unfortunately, however, this classroom setting is marred by a number of difficulties which hinder the process of learning in various ways. Limited time is assigned for imparting education with regard to each subject and theme and teachers are strictly expected to comply with the schedules allotted to them. In such a context, it may become extremely disconcerting for both teachers and students to establish the necessary levels of interaction and association. Teachers may find it cumbersome and challenging to accommodate the doubts, concerns, and queries of students within the limited time in the classroom. Students’ questions and doubts may go unresolved, which may have an adverse effect on their conceptual clarity and reading comprehension abilities and consequently affect their performance in board examinations.

Q.1

$\begin{array}{l} Prove that : \\ \sin^{2} \frac{π}{6} + \cos^{2} \frac{π}{3} - \tan^{2} \frac{π}{4} = - \frac{1}{2} \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \sin^{2} \frac{π}{6} + \cos^{2} \frac{π}{3} - \tan^{2} \frac{π}{4} = {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} - {(1)}^{2} \\ = \frac{1}{4} + \frac{1}{4} - 1 \\ = \frac{1 + 1 - 4}{4} \\ = - \frac{2}{4} \\ = - \frac{1}{2} \\ = R . H . S . Hence proved. \end{array}$

Q.2

$\begin{array}{l} Prove that : \\ 2 \sin^{2} (\frac{π}{6}) + {cosec}^{2} (\frac{7 π}{6}) \cos^{2} (\frac{π}{3}) = \frac{3}{2} \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ 2 \sin^{2} \frac{π}{6} + {cosec}^{2} \frac{7 π}{6} \cos^{2} \frac{π}{3} = 2 {(\frac{1}{2})}^{2} + {cosec}^{2} (π + \frac{π}{6}) \times {(\frac{1}{2})}^{2} \\ = 2 \times \frac{1}{4} + \frac{1}{4} \times {cosec}^{2} (\frac{π}{6}) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{1}{2} + \frac{1}{4} \times {(2)}^{2} \\ = \frac{1}{2} + 1 \\ = \frac{3}{2} = R . H . S . Hence proved. \end{array}$

Q.3

$\begin{array}{l} Prove that : \\ \cot^{2} \frac{π}{6} + cosec \frac{5 π}{3} + 3 \tan^{2} \frac{π}{6} = 6 \end{array}$

Ans.

$\begin{array}{l} L . H . S : \\ \cot^{2} \frac{π}{6} + cosec \frac{5 π}{6} + 3 \tan^{2} \frac{π}{6} = {(\sqrt{3})}^{2} + cosec (π - \frac{π}{6}) + 3 {(\frac{1}{\sqrt{3}})}^{2} \\ = 3 + cosec (\frac{π}{6}) + 3 \times \frac{1}{3} \\ [∵ cosec (π - x) = cosec x] \\ = 3 + 2 + 1 \\ = 6 = R . H . S . Hence proved. \end{array}$

Q.4

$\begin{array}{l} Prove that : \\ 2 \sin^{2} (\frac{3 π}{4}) + 2 \cos^{2} (\frac{π}{4}) + 2 \sec^{2} (\frac{π}{3}) = 10 \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ 2 \sin^{2} (\frac{3 π}{4}) + 2 \cos^{2} (\frac{π}{4}) + 2 \sec^{2} (\frac{π}{3}) = 2 \sin^{2} (π - \frac{π}{4}) + 2 {(\frac{1}{\sqrt{2}})}^{2} + 2 {(2)}^{2} \end{array}$

$\begin{array}{l} = 2 \sin^{2} (\frac{π}{4}) + 2 (\frac{1}{2}) + 8 \\ = 2 {(\frac{1}{\sqrt{2}})}^{2} + 1 + 8 \\ = 2 (\frac{1}{2}) + 1 + 8 \\ = 1 + 1 + 8 \\ = 10 = R . H . S . \end{array}$

Q.5 Find the value of: (i) sin75° (ii) tan15°

Ans.

$\begin{array}{l} (i) \sin 75 ° = \sin (45 ° + 30 °) \\ = \sin 45 ° \cos 30 ° + \sin 30 ° \cos 45 ° \\ [∵ \sin (x + y) = \sin x \cos y + \sin y \cos x] \\ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} \\ = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ (ii) \tan 15 ° = \tan (45 ° - 30 °) \\ = \frac{\tan 45 ° - \tan 30 °}{1 + \tan 45 ° \tan 30 °} \\ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} \end{array}$

$\begin{array}{l} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ = \frac{{(\sqrt{3} - 1)}^{2}}{3 - 1} \\ = \frac{3 - 2 \sqrt{3} + 1}{2} \\ \tan 15 ° = \frac{4 - 2 \sqrt{3}}{2} \\ = 2 - \sqrt{3} \end{array}$

Q.6 $\cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y) - \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y) = \sin (x + y)$

Ans.

$\begin{array}{l} Since, cosA cosB - sinA sinB = \cos (A + B) \\ L . H . S . = \\ \cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y) - \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y) \\ = \cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)} \\ = \cos (\frac{π}{2} - x - y) \\ = \cos {\frac{π}{2} - (x + y)} \\ = \sin (x + y) [∵ \cos (\frac{π}{2} - θ) = \sin θ] \\ = R . H . S . Hence proved. \end{array}$

Q.7

$\begin{array}{l} Prove the following : \\ \frac{\tan (\frac{π}{4} + x)}{\tan (\frac{π}{4} - x)} = {(\frac{1 + \tan x}{1 - \tan x})}^{2} \end{array}$

Ans.

$\begin{array}{l} L . H . S : \\ \frac{\tan (\frac{π}{4} + x)}{\tan (\frac{π}{4} - x)} = \frac{(\frac{\tan \frac{π}{4} + \tan x}{1 - \tan \frac{π}{4} \tan x})}{(\frac{\tan \frac{π}{4} - \tan x}{1 + \tan \frac{π}{4} \tan x})} [\begin{array}{l} ∵ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \end{array}] \\ = \frac{(\frac{1 + \tan x}{1 - 1 . \tan x})}{(\frac{1 - \tan x}{1 + 1 . \tan x})} \\ = (\frac{1 + \tan x}{1 - \tan x}) \times (\frac{1 + \tan x}{1 - \tan x}) \\ = {(\frac{1 + \tan x}{1 - \tan x})}^{2} \\ = R . H . S . Hence proved. \end{array}$

Q.8

$\begin{array}{l} Prove the following : \\ \frac{\cos (π + x) \cos (- x)}{\sin (π - x) \cos (\frac{π}{2} + x)} = \cot^{2} x \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\cos (π + x) \cos (- x)}{\sin (π - x) \cos (\frac{π}{2} + x)} = \frac{- \cos x \cos x}{\sin x \times - \sin x} [\begin{array}{l} ∵ \cos (π + x) = - cosx \\ \cos (- x) = cosx \\ \sin (π - x) = sinx \\ \cos (\frac{π}{2} + x) = - sinx \end{array}] \\ = \cot^{2} x \\ = R . H . S . Hence proved. \end{array}$

Q.9 $\cos (\frac{3 π}{2} + x) \cos (2 π + x) [\cot (\frac{3 π}{2} - x) + \cot (2 π + x)] = 1$

Ans.

$\begin{array}{l} L . H . S . = \cos (\frac{3 π}{2} + x) \cos (2 π + x) [\cot (\frac{3 π}{2} - x) + \cot (2 π + x)] \\ = \sin x . \cos x [\tan x + \cot x] \\ = \sin x . \cos x [\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}] \\ = \sin x . \cos x [\frac{\sin^{2} x + \cos^{2} x}{\sin x . \cos x}] \end{array}$

$\begin{array}{l} = \sin^{2} x + \cos^{2} x \\ = 1 = R . H . S . \end{array}$

Q.10 sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x

Ans.

$\begin{array}{l} L . H . S . : sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x \\ = cos (n + 1) x cos (n + 2) x + sin (n + 1) x sin (n + 2) x \\ = cos {(n + 1) x - (n + 2) x} \\ [∵ cosA cosB - sinA sinB = \cos (A + B)] \\ = cos (nx + x - nx - 2 x) \\ = cos (- x) [∵ \cos (- A) = cosA] \\ = \cos x = R . H . S . \end{array}$

Q.11 $\cos (\frac{3 π}{4} + x) - \cos (\frac{3 π}{4} - x) = - \sqrt{2} \sin x$

Ans.

$\begin{array}{l} L . H . S . : \\ \cos (\frac{3 π}{4} + x) - \cos (\frac{3 π}{4} - x) \\ = - 2 \sin (\frac{(\frac{3 π}{4} + x) + (\frac{3 π}{4} - x)}{2}) \sin (\frac{(\frac{3 π}{4} + x) - (\frac{3 π}{4} - x)}{2}) \\ = - 2 \sin (\frac{2 (\frac{3 π}{4})}{2}) \sin (\frac{2 x}{2}) \end{array}$

$\begin{array}{l} = - 2 \sin (π - \frac{π}{4}) \sin x \\ = - 2 \sin (π - \frac{π}{4}) \sin x \\ = - 2 \sin \frac{π}{4} \times \sin x \\ = - 2 \times \frac{1}{\sqrt{2}} \times \sin x \\ = - \sqrt{2} \sin x = R . H . S . \end{array}$

Q.12 Prove that:
sin² 6x – sin² 4x = sin 2x sin 10x

Ans.

$\begin{array}{l} L . H . S . = {sin}^{2} 6x - {sin}^{2} 4x \\ = (\sin 6 x + \sin 4 x) (\sin 6 x - \sin 4 x) [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = {2 \sin (\frac{6 x + 4 x}{2}) \cos (\frac{6 x - 4 x}{2})} {2 \cos (\frac{6 x + 4 x}{2}) \sin (\frac{6 x - 4 x}{2})} \\ = {2 \sin 5 x \cos x} {2 \cos 5 x \sin x} \\ = (2 \sin x \cos x) (2 \sin 5 x \cos 5 x) \\ = \sin 2 x . \sin 10 x [∵ \sin 2 x = 2 \sin x \cos x] \\ = R . H . S . Hence proved. \end{array}$

Q.13 Prove that:
cos²2x – cos²6x = sin 4x sin 8x

Ans.

$\begin{array}{l} L . H . S . = \cos^{2} 2x - \cos^{2} 6x \\ = (\cos 2 x + \cos 6 x) (\cos 2 x - \cos 6 x) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = {2 \cos (\frac{2 x + 6 x}{2}) \cos (\frac{2 x - 6 x}{2})} {- 2 \sin (\frac{2 x - 6 x}{2}) \sin (\frac{2 x + 6 x}{2})} \\ = {2 \cos 4 x \cos (- 2 x)} {- 2 \sin (- 2 x) \sin 4 x} \\ = (2 \sin 2 x \cos 2 x) (2 \sin 4 x \cos 4 x) \\ = \sin 4 x . \sin 8 x [∵ \sin 2 x = 2 \sin x \cos x] \\ = R . H . S . Hence proved. \end{array}$

Q.14 Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos²x sin 4x

Ans.

$\begin{array}{l} sin 2x + 2 sin 4x + sin 6x = 2 sin 4x + sin 6x + sin 2x \\ = 2 sin 4x + 2 \sin (\frac{6 x + 2 x}{2}) \cos (\frac{6 x - 2 x}{2}) \\ = 2 sin 4x + 2 \sin 4 x \cos 2 x \\ = 2 sin 4x (1 + \cos 2 x) \\ = 2 sin 4x (2 \cos^{2} x) \\ = 4 \cos^{2} x sin 4x \\ = R . H . S . \end{array}$

Q.15 Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Ans.

$\begin{array}{l} We have, cot 4x (sin 5x + sin 3x) = cot x (sin 5x - sin 3x) \\ The given equation can be written as: \\ \frac{sin 5x + sin 3x}{sin 5x - sin 3x} = \frac{cot x}{cot 4x} \end{array}$

$\begin{array}{l} L . H . S : \frac{sin 5x + sin 3x}{sin 5x - sin 3x} = \frac{2 \sin (\frac{5 x + 3 x}{2}) \cos (\frac{5 x - 3 x}{2})}{2 cos \frac{5 x + 3 x}{2} sin \frac{5 x - 3 x}{2}} \\ = \frac{\sin 4 x \cos x}{\cos 4 x sin x} \\ = \frac{(\frac{\cos x}{\sin x})}{(\frac{\cos 4 x}{\sin 4 x})} \\ = \frac{\cot x}{\cot 4 x} \\ = R . H . S . Hence proved. \end{array}$

Q.16

$\begin{array}{l} Prove that : \\ \frac{\cos 9 x - \cos 5 x}{\sin 17 x - \sin 3 x} = - \frac{\sin 2 x}{\cos 10 x} \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\cos 9 x - \cos 5 x}{\sin 17 x - \sin 3 x} = \frac{- 2 \sin (\frac{9 x + 5 x}{2}) \sin (\frac{9 x - 5 x}{2})}{2 \cos (\frac{17 x + 3 x}{2}) \sin (\frac{17 x - 3 x}{2})} \\ [\begin{array}{l} ∵ cosA - cosB = - 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \\ sinA - sinB = - 2 \cos (\frac{A - B}{2}) \sin (\frac{A - B}{2}) \end{array}] \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = - \frac{\sin (7 x) \sin (2 x)}{\cos (10 x) \sin (7 x)} \\ = - \frac{\sin 2 x}{\cos 10 x} = R . H . S . Hence proved. \end{array}$

Q.17

$\begin{array}{l} Prove that : \\ \frac{\sin 5 x + \sin 3 x}{\cos 5 x + \cos 3 x} = \tan 4 x \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\sin 5 x + \sin 3 x}{\cos 5 x + \cos 3 x} = \frac{2 \sin (\frac{5 x + 3 x}{2}) \cos (\frac{5 x - 3 x}{2})}{2 \cos (\frac{5 x + 3 x}{2}) \cos (\frac{5 x - 3 x}{2})} \\ = \frac{\sin 4 xcosx}{\cos 4 xcosx} \\ = \tan 4 x \\ = R . H . S . Hence proved. \end{array}$

Q.18

$\begin{array}{l} Prove that : \\ \frac{sinx - siny}{cosx + cosy} = \tan (\frac{x - y}{2}) \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{sinx - siny}{cosx + cosy} = \frac{2 \cos (\frac{x + y}{2}) \sin (\frac{x - y}{2})}{2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2})} \\ = \frac{\sin (\frac{x - y}{2})}{\cos (\frac{x - y}{2})} \\ = \tan (\frac{x - y}{2}) \\ = R . H . S . Hence proved. \end{array}$

Q.19

$\begin{array}{l} Prove that : \\ \frac{sinx + \sin 3 x}{cosx + \cos 3 x} = \tan 2x \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{sinx + \sin 3 x}{cosx + \cos 3 x} = \frac{\sin 3 x + sinx}{\cos 3 x + cosx} \\ = \frac{2 \sin (\frac{3 x + x}{2}) \cos (\frac{3 x - x}{2})}{2 \cos (\frac{3 x + x}{2}) \cos (\frac{3 x - x}{2})} \\ = \frac{\sin 2 x}{\cos 2 x} \\ = \tan 2 x \\ = R . H . S . Hence proved. \end{array}$

Q.20

$\begin{array}{l} Prove that : \\ \frac{\sin x - \sin 3 x}{\sin^{2} x - \cos^{2} x} = 2 \sin x \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\sin x - \sin 3 x}{\sin^{2} x - \cos^{2} x} = \frac{- (\sin 3 x - \sin x)}{- (\cos^{2} x - \sin^{2} x)} \\ = \frac{2 \cos (\frac{3 x + x}{2}) \sin (\frac{3 x - x}{2})}{\cos 2 x} \\ = \frac{2 \cos 2 x \sin x}{\cos 2 x} \\ = 2 \sin x \\ = R . H . S . \end{array}$

Q.21

$\begin{array}{l} Prove that : \\ \frac{\cos 4 x + \cos 3 x + \cos 2 x}{\sin 4 x + \sin 3 x + \sin 2 x} = \cot 3 x \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\cos 4 x + \cos 3 x + \cos 2 x}{\sin 4 x + \sin 3 x + \sin 2 x} = \frac{\cos 4 x + \cos 2 x + \cos 3 x}{\sin 4 x + \sin 2 x + \sin 3 x} \\ = \frac{2 \cos (\frac{4 x + 2 x}{2}) \cos (\frac{4 x - 2 x}{2}) + \cos 3 x}{2 \sin (\frac{4 x + 2 x}{2}) \cos (\frac{4 x - 2 x}{2}) + \sin 3 x} \\ = \frac{2 \cos 3 x \cos x + \cos 3 x}{2 \sin 3 x \cos x + \sin 3 x} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{\cos 3 x (2 \cos x + 1)}{\sin 3 x (2 \cos x + 1)} \\ \frac{\cos 4 x + \cos 3 x + \cos 2 x}{\sin 4 x + \sin 3 x + \sin 2 x} = \cot 3 x = R . H . S . Hence proved. \end{array}$

Q.22 Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Ans.

$\begin{array}{l} ∵ \cot 3 x = \cot (2 x + x) \\ = \frac{\cot 2 x \cot x - 1}{\cot 2 x + \cot x} \\ \cot 3 x (\cot 2 x + \cot x) = \cot 2 x \cot x - 1 \\ \cot 3 x \cot 2 x + \cot 3 x \cot x = \cot 2 x \cot x - 1 \\ 1 = \cot x \cot 2 x - \cot 2 x \cot 3 x - \cot 3 x \cot x \\ or \cot x \cot 2 x - \cot 2 x \cot 3 x - \cot 3 x \cot x = 1 . Hence proved. \end{array}$

Q.23

$\begin{array}{l} Prove that : \\ \tan 4 x = \frac{4 \tan x (1 - \tan^{2} x)}{1 - 6 \tan^{2} x + \tan^{4} x} \end{array}$

Ans.

$\begin{array}{l} L . H . S : \\ \tan 4 x = \tan 2 (2 x) \\ = \frac{2 \tan 2 x}{1 - \tan^{2} 2 x} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{2 (\frac{2 \tan x}{1 - \tan^{2} x})}{1 - {(\frac{2 \tan x}{1 - \tan^{2} x})}^{2}} \\ = \frac{(\frac{4 \tan x}{1 - \tan^{2} x})}{{\frac{{(1 - \tan^{2} x)}^{2} - 4 \tan^{2} x}{{(1 - \tan^{2} x)}^{2}}}} \\ = \frac{4 \tan x}{1 - \tan^{2} x} \times \frac{{(1 - \tan^{2} x)}^{2}}{1 - 2 \tan^{2} x + \tan^{4} x - 4 \tan^{2} x} \\ = \frac{4 \tan x (1 - \tan^{2} x)}{1 - 6 \tan^{2} x + \tan^{4} x} = R . H . S . \\ Hence proved. \end{array}$

Q.24 Prove that:
cos 4x = 1 – 8sin² x cos² x

Ans.

$\begin{array}{l} L . H . S . : \cos 4 x = \cos 2 (2 x) \\ = 1 - 2 \sin^{2} (2 x) [∵ \cos 2 x = 1 - \sin^{2} x] \\ = 1 - 2 {(2 \sin x \cos x)}^{2} [∵ \sin 2 x = 2 \sin x \cos x] \\ = 1 - 2 (4 \sin^{2} x \cos^{2} x) \\ = 1 - 8 \sin^{2} x \cos^{2} x \\ = R . H . S . \end{array}$

Q.25 Prove that:
cos 6x = 32 cos⁶ x – 48cos⁴ x + 18 cos² x – 1

Ans.

$\begin{array}{l} L . H . S . = cos 6x \\ = \cos 3 (2 x) \\ = 4 \cos^{3} 2 x - 3 \cos 2 x \\ = 4 {(2 \cos^{2} x - 1)}^{3} - 3 (2 \cos^{2} x - 1) \\ = 4 (8 \cos^{6} x - 12 \cos^{4} x + 6 \cos^{2} x - 1) - 6 \cos^{2} x + 3 \\ = 32 \cos^{6} x - 48 \cos^{4} x + 24 \cos^{2} x - 4 - 6 \cos^{2} x + 3 \\ = 32 \cos^{6} x - 24 \cos^{4} x + 18 \cos^{2} x - 1 \\ = R . H . S . Hence proved. \end{array}$

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