NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry - Future Study Point

NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry

NCERT solutions class 10 maths exercise 8.3

NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry

NCERT solutions class 10 maths exercise 8.3

NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry are presented here by Future Study Pointed free of cost. These NCERT Solutions of Exercise 8.3-Introduction to Trigonometry will boost your preparation of the forthcoming exams and in doing your homework and class assignments.You can study here all the exercises of the chapter 8-Introduction to Trigonometry.All questions are solved by an expert of maths as per the CBSE norms prescribed for class 10 standard.

The NCERT Solutions for class 10 maths are the most important study material for class 10 in clearing their doubts on maths. The marks in mathematics play an important role in increasing your aggregate percentage in the result of the CBSE board exam, so every student required special attention in mathematics.

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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 8.3 of Chapter 8 Introduction to Trigonometry

Q1.Evaluate

Ans.

(i) sin 18°/cos 72°

Since sin θ = cos (90°-θ)

⇒sin18°=cos(90°-18°)

Substituting the value of sin 18°

cos(90°-18°)/cos 72°

cos 72°/cos 72° = 1

(ii) tan 26°/cot 64°

Since tan θ = cot (90°-θ)

⇒tan26°=cot(90°-26°)

Substituting the value of tan26°

cot(90°-26°)/cot 72°

cot 72°/cot72° = 1

(iii) cos 48° – sin 42°

Since cos θ = sin (90°-θ)

⇒cos 48°=sin(90°-48°)

Substituting the value of sin 18°

sin(90°-48°)° – sin 42°

sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

Since cosec θ = sec (90°-θ)

⇒cosec 31°=sec(90°-31°)

Substituting the value of cosec 31°

sec(90°-31°) – sec 59°

sec 59° – sec 59° = 0

Q2.Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Ans.

(i) tan 48° tan 23° tan 42° tan 67° = 1

Since tan θ = cot (90°-θ)

⇒tan 48°=cot(90°-48°) and ⇒tan 43°=cot(90°-43°)

Substituting the value of tan 48° and tan 43°

cot(90°-48°) cot(90°-43°) tan 42° tan 67°

cot42° cot67° tan 42° tan 67°

Substituting cot42° =1/tan42° and cot67° =1/tan67°

(1/tan42°) (1/tan 67° )tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Since cos θ = sin (90°-θ)

⇒cos 38°=sin(90°-38°) and ⇒sin38°=cos(90°-38°)

Substituting the value of cos38° and sin38°

sin(90°-38°). cos 52°- cos(90°-38°).sin 52°

sin 52°.cos 52° – cos 52°sin 52° = 0

Q3.If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A

Ans. The given equation is

tan 2A = cot (A – 18°), where 2A is an acute angle

cot (90°-2A) = cot (A – 18°)

90°-2A = A – 18°

A + 2A =90°+ 18°=108°

3A = 108°

A = 36°

Q4.If tan A = cot B, prove that A + B = 90°.

Ans. The given equation is

tan A = cot B

cot (90°-  A) =  cot B

90°-  A =  B

A  + B =90°,Hence proved

Q5.If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A

Ans.The given equation is

sec 4A = cosec (A – 20°),where 4A is an acute angle

cosec (90°- 4A) = cosec (A – 20°)

90°- 4A = A – 20°

A+4A = 90° +20°=110°

5A = 110°

A = 22°

Q6.If A, B and C are interior angles of a triangle ABC, then show that:

Ans.According to angle sum property of triangle

A + B + C= 180°

B + C = 180°- A

Dividing by 2

Q7.Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Ans. The given expression is

sin 61° + cos 75

cos(90° – 61°) + sin(90°-75°)

cos 29° +sin 15°

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
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
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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
Chapter 8-Trigonometry

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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

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NCERT solutions of 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 of 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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