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# 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

(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

Please see this video and subsribe ,like and share it

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

### NCERT Solutions of class 9 science

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### NCERT Solutions of class 10 maths

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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

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### 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

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