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
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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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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 |
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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 |
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Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
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