**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.

**NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise**

**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 .**

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**NCERT Solutions for 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**

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**NCERT solutions for class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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