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 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 System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
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Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 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-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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NCERT solutions of 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