**Class 11 th CBSE NCERT Maths Solution 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.

DOWNLOAD class 11 th maths PDF **chapter 3 trigonometry** A TO Z** NCERT solutions** from the following link

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**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 NCERT solutions of Physics and Chemistry**

**Chapter 1 – Some concepts of the chemistry**

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

**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 .**

Study exercise 3.1, 3.2, 3.3 and 3.4 , click the links.

Class Xl CBSE Maths(NCERT) Solution of Trigonometry Chapter 3.1 and 3.2

CLASS 11 th CBSE MATHS NCERT SOLUTIONS EXERCISE 3.3

CLASS 11 th CBSE MATHS NCERT SOLUTIONS EXERCISE 3.4

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