NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry - Future Study Point

NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry

Exercise 8.4 class 10 maths

NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry are the best study inputs for the preparations of trigonometry questions in the CBSE board exam of 10 class since maximum questions are asked from this exercise 8.4 -Introduction to Trigonometry. All questions of Exercise 8.4 are solved by an expert of maths as per the CBSE norms.  NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry are explained here beautifully so that every students of 10 standards could understand properly. You can study here science and maths notes, e-books of science and maths and notes for the carrier, preparation of competitive entrance exams.

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NCERT Solutions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Exercise 8.1- Introduction to Trigonometry

Exercise 8.2 -Introduction to Trigonometry

Exercise 8.3-Introduction to Trigonometry

Exercise 8.4-Introduction to Trigonometry

NCERT solutions of Important Questions-Introduction to Trigonometry

Latest Sample paper Class 10 maths for Term 1 2021 CBSE board

NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry

Q1.Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

Ans.sin A, sec A and tan A can be transformed into the terms of cot A as follows.

Transforming the sin A into cosec A

sin A = 1/cosec A

Applying the identity

cosec² A =  1 + cot² A

cosec A = √( 1 + cot² A)

Substituting the value of cosec A

sec A = √( 1 + tan²A)

tan A = 1/cot A

Q2.Write all the other trigonometric ratios of ∠A in terms of sec A.

Ans.

sin A = √( 1 – cos²A)

cos A =1/sec A

tan A = √(sec²A -1)

cot A = 1/tan A

cosec A = 1/sin A

Q3. Evaluate

(ii) sin 25°cos 65°+ cos 25°sin 65°

Ans.

Since sin 63° = cos (90° – 63°) and cos 17° = sin(90° – 17°)

Substituting the value of sin63° and cos 17°

= 1/1 = 1

(ii) sin 25°cos 65°+ cos 25°sin 65°

Since sin 25°= cos( 90° – 25°) and cos 25° = sin(90° – 25°)

Substituting the value of sin 25° and cos 25°

cos( 90° – 25°)cos 65°+ sin(90° – 25°)sin 65°

cos 65°.cos 65° +sin 65°.sin 65°

cos² 65° + sin² 65° = 1

Q4.Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A = ……
(A) 1
(B) 9
(C) 8
(D) 0

Ans. (i) 9 sec² A – 9 tan² A

=9(1 + tan²A) – 9 tan² A

= 9 + 9 tan²A – 9 tan² A

=9

Hence the answer is (B) 9

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

(A) 0

(B) 1

(C) 2

(D) -1

Ans.(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Coverting all trigonometric function in tems of sinθ and cosθ

( 1 + sinθ/cos θ  + 1/cosθ)( 1 + cosθ/sinθ  – 1/sinθ)

(cos θ + sinθ + 1)/cosθ(sinθ + cosθ – 1)/sinθ

[(cos θ + sinθ)² – 1²]/cos θ sinθ

[cos²θ + sin²θ+ 2cos θ sinθ-1]/cos θ sinθ

[1 + 2cos θ sinθ-1]/cos θ sinθ = 2cos θ sinθ/cos θ sinθ =2

Hence the answer is

(C) 2

(A) sec²A

(B)-1

(C)cot²A

(D)tan²A

Ans.

Applying the identies 1 + tan²A = sec²A and 1 + cot²A = cosec²A

Hence the answer is (D) tan²A

Q5.Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosecθ – cotθ)² =(1-cosθ)/(1+cosθ)

Ans.The given identy is

Taking LHS

Converting the trigonometric function into sin and cos

Taking LHS

Taking LHS

Multiplying the numerator and denominator of second fraction by -1

Applying the identity a³ – b³= (a – b)(a² +b²+ab)

Since sin²θ + cos²θ = 1

See the video -class 10 maths exercise 8.4 question 5 part iv

Ans. Taking LHS

simplifying  LHS

 

Multiplying the deniminator and numerator by (1-cosθ)

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Ans. Taking LHS

Dividing numerator and denominator by sin A

Rearranging the terms of numerator and denominator

Multiplying the numerator and denominator by the conjugate of cotA -cosecA(i.e cotA +cosecA)

Since cot²A = cosec²A – 1

 

Ans. Taking the LHS

Multiplying the denominator and numerator by the conjugate of denominator (1 + sinA)

We have to prove

Taking LHS

Taking sinθ and cosθ from the numerator and denominator respectively

Since sinθ/cosθ = tanθ and sin²θ = 1 – cos²θ

Ans. Taking LHS

Applying the identity (a + b)² = a² + b² + 2ab

=sin²A + cosec²A + 2sinA.cosecA +cos²A + sec²A + 2cosA.secA

Arranging the terms and applying the identities 1 + cot²A = cosec²A and 1+tan²A = sec²A

=sin²A + cos²A + 1 + cot²A + 1+tan²A +2sinA.1/sinA +2cosA.1/secA

=1 +1 +1 +2+2 + tan²A + cot²A

=7 + tan²A + cot²A=RHS

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Ans.Taking LHS

Transforming the LHS into the terms of sin and cos

Dividing and multiplying it by (tanA + cot A)

 

Ans.Taking

converting the terms into sin and cos

Taking

Hence it is verified that

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

NCERT Solutions of 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

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Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
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Chapter3- Atoms and MoleculesChapter 11- Work and Energy
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Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
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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
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Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
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Chapter 4- Carbon and its CompoundsChapter 12- Electricity
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Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

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Chapter 1-SetsChapter 9-Sequences and Series
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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
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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
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