NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise - Future Study Point

NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

miscellaneous ex.chapter 3 class 11 ncert solutions

NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

The NCERT  maths solutions of chapter 3  miscellaneous exercise containing 10 questions is the extract of the lesson ‘Trigonometry’ . The questions of this exercise are very important for the purpose of your academic exams and entrance exams of engineering.The solutions are explained discretely in order to clear the doubts generally faced by  11 th class students .All questions are explained readily by the teacher expertise in maths. We hope that these solutions by Future Study Point will help you in your preparations of the exam. After you go through each solution write your comment in the form of suggestions, doubts and your requirement.

miscellaneous ex.chapter 3 class 11 ncert solutions

NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

Exercise 3.1 and Exercise 3.2

Exercise 3.3

 Exercise 3.4

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

Class 11 th CBSE NCERT Maths Solution of Chapter 3 Miscellaneous Exercise

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Q1. Prove that.

Answer.

L.H.S

cosA + cosB =2cos(A+B)/2.cos(A–B)/2

As we know cos(–x) = cosx

= 0 = R.H.S

Q2. Prove that

(sin3x + sinx)sinx  + (cos3x – cosx)cosx = 0

∵sinA + sinB = 2sin(A+B)/2.cos(A–B)/2

∵cosA –cosB = –2sin(A+B)/2.sin(A–B)/2

=2sin2x.cosx.sinx – 2sin2x.sinx.cosx  = 0= R.H.S

Q3.Prove that

Answer.

L.H.S

=cos²x + cos²y + 2cosx.cosy + sin²x + sin²y –2sinx.siny

=(cos²x + sin²x  )+ (cos²y + sin²y) + 2(cosx.cosy – sinx.siny)

=1 + 1 + 2(cosx.cosy – sinx.siny)

=2 + 2(cosx.cosy – sinx.siny)

∵cos(A +B) = cosA.cosB – sinA.sinB

=2 + 2cos(x + y)

=2[1 + cos(x + y)]

∵  cos2A = 2cos²A – 1

⇒1+cos2A  = 2cos²A

Q4.Prove that

Answer.

L.H.S     (cosx – cosy)² + (sinx – siny)²

=cos²x + cos²y – 2cosx.cosy + sin²x + sin²y –2sinx.siny

=(cos²x  +  sin²x) + (cos²y  + sin²y) –2(cosx.cosy + sinx.siny)

= 1 + 1 – 2cos(x – y)  [∵ cos(A – B) = cosA.cosB + sinA.sinB]

= 2 – 2cos(x– y)

=2[1 – cos(x –y)]

∵ cos2A = 1 – 2sin²A⇒ 1 – cos2A  = 2sin²A

= R.H.S

Q5. Prove that  sinx + sin3x + sin5x + sin7x = 4cosx.cos2x.sin4x

Answer.

L.H.S

=sinx + sin3x + sin5x + sin7x

= (sinx + sin3x )+ (sin5x + sin7x)

∵ sinA + sinB = 2sin(A + B)/2 .cos(A –B)/2

=2sin2x.cos(–x)  + 2sin6x.cos(–x)

∵ cos(–x) = cosx

=2sin2x.cosx + 2sin6x.cosx

=2cosx(sin2x + sin6x)

∵ sinA + sinB = 2sin(A + B)/2 .cos(A –B)/2

= 4cosx.sin4x.cos(–2x)

∵ cos(–θ) = cosθ

= 4cosx.cos2x.sin4x = R.H.S

Q6. Prove that

Answer.

L.H.S

= tan6x = R.H.S

 

Q7. Prove that

Answer. L.H.S = sin3x + (sin2x – sinx)

∵sin2A = 2sinA.cosA

Answer.

Since x is in 2nd quadrant, so the value of x would be estimated as follows

Dividing all the terms of this equation of inequality

The value of will lye in l quadrant.

We are given that

 

Placing the value of tanx =(–4/5)

As we have shown above that x/2 lyes in 1st quadrant,tan(x/2) will be positive

x/2 lies in 1 st quadrant so sec(x/2) =√5

sinx/2 = ±√(1–cos²x/2)

x/2 lies in 1st quadrant so sin(x/2) will be positive

Answer.  x is in lll quadrant, so its value would be estimated as follows

π < x < 3π/2, dividing this eq. by 2,we shall have

π/2 < x/2<3π/4, which reveals that x/2  lyes in ll quadrant .

∵cos2x = 2cos²x –1

As we have shown above that x/2 lies in ll quadrant and it is known to us ,the value of cos  is negative in 2nd quadrant.

As we have shown above that x/2 lies in ll quadrant and it is known to us ,the value of sin x/2 is positive in 2nd quadrant.

Answer.

Since ,x is in ll quadrant ,its value would be estimated as follows

π/2 < x < π

⇒π/4 < x/2 < π/2, implies that  x/2 lies in l st quadrant, which shows that all sinx/2, cos x/2 and tan x/2 will lye in 1st quadrant.

x is in ll quadrant ,so cosx will be negative

cos2x = 2cos²x – 1

 

We have shown that x/2 lyes in l quadrant and value of cos is positive in l quadrant.

As it is already known to us that x/2 is  in l st quadrant and value of sin x/2 is postitive in lst quadrant.

Multiplying the denominator and numerator by the conjugate of denominator .

 

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

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NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

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NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

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