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

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

NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4 is going to guide you through the most crucial exercise in Chapter 8. The exercise generally addresses the problems involved in trigonometric identities, a stepping stone towards the intricate concepts in later classes. The exercise focuses on the connection between trigonometric identities and various trigonometric ratios. We will continue with this important concept, which will benefit you in your board exams as well as in later math courses.

1.0Download NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4 : Free PDF

NCERT Solutions Class 10 maths Chapter 8 Introduction to Trigonometry Exercise 8.4

2.0Introduction to Trigonometric Identities: 

Trigonometric identities are formulas that include trigonometric ratios (sine, cosine, tangent, etc.) of an angle and hold for all the values of the concerned angles.These are basic mathematical formulas that help simplify expressions and work out problems that require understanding the interrelations among different trigonometric functions. In this exercise, we shall be studying some of the fundamental trigonometric identities and their proofs.

3.0Exercise 8.4 Overview: Key Concepts

The exercise mentions three main trigonometric identities for the right-angled triangle. These identities will continuously be used in the exercise to solve various problems of trigonometry. 

The Pythagorean Identity: 

One of the most significant and basic trigonometric identities is the Pythagorean identity used in exercise 8.4. 

This identity comes from the Pythagorean theorem when applied to a right-angled triangle. According to the theorem,  the sides of a right-angled triangle share the following relation: 

AB2+BC2=AC2

Here, AB, BC, and AC represent the sides of the triangle, with AC being the hypotenuse. 

After dividing both sides of the above equation by AC2, it can be re-written as the following Pythagorean identity of trigonometry: 

Cos2A+Sin2A=1

In this identity, A is the angle formed opposite to the perpendicular of the triangle such that, 0°A90°. The condition mentioned here holds true for all angles A. 

The Secant-Tangent Identity: 

This identity connects the secant and tangent functions. It states the result of the Pythagorean theorem by dividing both sides of the formula by the square of the right triangle's adjacent side. The identity states that the square of the secant of an angle equals one plus the square of the tangent of the same angle. This identity comes in handy when simplifying expressions and solving equations containing these two functions in exercise.

1+tan2A=Sec2A

In this identity, 0°A<90°. 

The Cosecant-Cotangent Identity: 

The cosecant-cotangent identity relates the cosecant and cotangent functions. It is also obtained from the Pythagorean theorem, but this time by dividing the formula by the square of the opposite side. The identity demonstrates that the square of the cosecant of an angle equals one plus the square of the cotangent of the same angle. It assists in representing one trigonometric function in terms of another and finds applications in numerous trigonometric simplifications. The identity can mathematically be written as: 

1+Cot2A=Cosec2A

The identity follows the rule for all values of A as “0°<A90°” 

Expressing different trigonometric ratios in terms of another ratio: 

The trigonometric identities help simplify the process of writing different trigonometric ratios in terms of another ratio. For example, different ratios of trigonometry can be written in terms of “Sin” by interchanging and doing arithmetic operations on the three identities mentioned earlier. This forms a basic concept for solving different questions of exercise 8.4. 

Mastering the three trigonometric identities will not only sharpen your problem-solving skills but also lay a strong foundation for your future in mathematics. This is why you should keep practicing these identities with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4. 

4.0NCERT Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4 : Detailed Solutions

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

Solution:

We have cosec² A - cot² A = 1

=> cosec² A = 1 + cot² A

=> (cosec A)² = cot² A + 1

=> (1/sin A)² = cot² A + 1

=> (sin A)² = 1/(cot² A + 1)

=> sin A = ± 1/√(cot² A + 1)

We reject the negative value of sin A for acute angle A. Therefore, sin A = 1/√(cot² A + 1).

Also, tan A = 1/cot A.

We have sec² A - tan² A = 1

=> sec² A = 1 + tan² A

=> sec² A = 1 + 1/cot² A = (cot² A + 1)/cot² A

=> sec A = √(cot² A + 1)/cot A


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

Solution:

(i) sin A = √(1 - cos² A)

= √(1 - 1/sec² A) = √(sec² A - 1)/sec A

(ii) cos A = 1/sec A

(iii) tan A = √(sec² A - 1)

(iv) cot A = 1/tan A = 1/√(sec² A - 1)

(v) cosec A = 1/sin A = sec A/√(sec² A - 1)


3. Evaluate:

(i) (sin² 63° + sin² 27°) / (cos² 17° + cos² 73°)

(ii) sin 25° cos 65° + cos 25° sin 65°

Solution:

(i) (sin² 63° + sin² 27°) / (cos² 17° + cos² 73°)

= {sin(90° - 27°)}² + sin² 27° / cos² 17° + {cos(90° - 17°)}²

= {cos 27°}² + sin² 27° / cos² 17° + {sin 17°}²

= (cos² 27° + sin² 27°) / (cos² 17° + sin² 17°) = 1/1 = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

= sin(90° - 65°) cos 65° + cos(90° - 65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos² 65° + sin² 65° = 1


4. Choose the correct option. Justify your choice:

(i) 9 sec² A - 9 tan² A =

(A) 1

(B) 9

(C) 8

(D) 0

(ii) (1 + tan θ + sec θ)(1 + cot θ - cosec θ) =

(A) 0

(B) 1

(C) 2

(D) -1

(iii) (sec A + tan A)(1 - sin A) =

(A) sec A

(B) sin A

(C) cosec A

(D) cos A

(iv) (1 + tan² A) / (1 + cot² A) =

(A) sec² A

(B) -1

(C) cot² A

(D) tan² A

Solution:

(i) Option (B)

9 sec² A - 9 tan² A = 9(sec² A - tan² A) = 9 × 1 = 9.

(ii) Option (C)

(1 + tan θ + sec θ)(1 + cot θ - cosec θ)

= {1 + sin θ/cos θ + 1/cos θ} × {1 + cos θ/sin θ - 1/sin θ}

= {(cos θ + sin θ + 1)/cos θ} × {(sin θ + cos θ - 1)/sin θ}

= {(cos θ + sin θ) + 1} × {(cos θ + sin θ) - 1} / (cos θ × sin θ)

= {(cos θ + sin θ)² - 1²} / (cos θ × sin θ)

{∵ (a + b)(a - b) = a² - b²}

= (cos² θ + sin² θ + 2 cos θ sin θ - 1) / (cos θ × sin θ)

= (1 + 2 cos θ sin θ - 1) / (cos θ sin θ) = 2.

(iii) Correct option is (D).

(sec A + tan A)(1 - sin A)

= sec A - sec A sin A + tan A - tan A sin A

= 1/cos A - sin A/cos A + sin A/cos A - sin² A/cos A

= (1 - sin² A) / cos A

= cos² A / cos A = cos A

(iv) Correct option is (D).

(1 + tan² A) / (1 + cot² A) = sec² A / cosec² A = tan² A


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

(i) (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ)

(ii) (cos A / (1 + sin A)) + ((1 + sin A) / cos A) = 2 sec A

(iii) (tan θ / (1 - cot θ)) + (cot θ / (1 - tan θ)) = 1 + sec θ cosec θ

(iv) (1 + sec A) / sec A = (sin² A) / (1 - cos A)

(v) (cos A - sin A + 1) / (cos A + sin A - 1) = cosec A + cot A, using the identity cosec² A = 1 + cot² A.

(vi) √((1 + sin A) / (1 - sin A)) = sec A + tan A

(vii) (sin θ - 2 sin³ θ) / (2 cos³ θ - cos θ) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

(ix) (cosec A - sin A)(sec A - cos A) = 1 / (tan A + cot A)

(x) ((1 + tan² A) / (1 + cot² A)) = ((1 - tan A) / (1 - cot A))² = tan² A

Solution:

(i) LHS = (cosec θ - cot θ)²

= (1/sin θ - cos θ/sin θ)² = ((1 - cos θ) / sin θ)²

= (1 - cos θ)² / sin² θ = (1 - cos θ)² / (1 - cos² θ)

= (1 - cos θ)² / ((1 - cos θ)(1 + cos θ)) = (1 - cos θ) / (1 + cos θ)

∴ LHS = RHS.

(ii) LHS = (cos A / (1 + sin A)) + ((1 + sin A) / cos A)

= (cos² A + (1 + sin A)²) / (cos A(1 + sin A))

= (cos² A + 1 + sin² A + 2 sin A) / (cos A(1 + sin A))

= (2(1 + sin A)) / (cos A(1 + sin A))

= 2 sec A = RHS.

(iii) LHS = (tan θ / (1 - cot θ)) + (cot θ / (1 - tan θ))

= ((sin θ / cos θ) / (1 - cos θ / sin θ)) + ((cos θ / sin θ) / (1 - sin θ / cos θ))

= ((sin θ / cos θ) / (1 - cos θ / sin θ)) + ((cos θ / sin θ) / (1 - sin θ / cos θ))

= (sin θ × sin θ) / (cos θ(sin θ - cos θ)) + (cos θ × cos θ) / (sin θ(cos θ - sin θ))

= (sin² θ) / (cos θ(sin θ - cos θ)) - (cos² θ) / (sin θ(sin θ - cos θ))

= (sin³ θ - cos³ θ) / (cos θ sin θ (sin θ - cos θ))

= ((sin θ - cos θ)(sin² θ + cos² θ + sin θ cos θ)) / (cos θ sin θ (sin θ - cos θ))

{∵ a³ - b³ = (a - b)(a² + b² + ab)}

= (sin² θ + cos² θ + sin θ cos θ) / (cos θ sin θ)

= (1 + sin θ cos θ) / (cos θ sin θ) = 1 / (cos θ sin θ) + 1

= 1 + (1 / cos θ)(1 / sin θ)

= 1 + sec θ cosec θ

∴ LHS = RHS.

(iv) LHS = (1 + sec A) / sec A = (1 + 1/cos A) / (1/cos A) = (cos A + 1) / 1 = 1 + cos A

RHS = (sin² A) / (1 - cos A) = (1 - cos² A) / (1 - cos A) = 1 + cos A

∴ LHS = RHS.

(v) LHS = (cos A - sin A + 1) / (cos A + sin A - 1)

= ((cos A / sin A) - (sin A / sin A) + (1 / sin A)) / ((cos A / sin A) + (sin A / sin A) - (1 / sin A))

(Dividing the numerator and denominator by sin A)

= (cot A - 1 + cosec A) / (cot A + 1 - cosec A) = ((cosec A + cot A) - 1) / (1 + cot A - cosec A)

= ((cosec A + cot A) - (cosec² A - cot² A)) / (1 + cot A - cosec A)

(∵ cosec² A = 1 + cot² A, i.e., cosec² A - cot² A = 1)

= ((cosec A + cot A) - (cosec A + cot A)(cosec A - cot A)) / (1 + cot A - cosec A)

{∵ (a + b)(a - b) = a² - b²}

= ((cosec A + cot A)(1 - (cosec A - cot A))) / (1 + cot A - cosec A)

= ((cosec A + cot A)(1 + cot A - cosec A)) / (1 + cot A - cosec A)

= cosec A + cot A = RHS.

(vi) LHS = √((1 + sin A) / (1 - sin A)) = √(((1 + sin A)(1 + sin A)) / ((1 - sin A)(1 + sin A)))

= √((1 + sin A)² / (1 - sin² A)) = √((1 + sin A)² / cos² A) = (1 + sin A) / cos A

= 1 / cos A + sin A / cos A = sec A + tan A

∴ LHS = RHS.

(vii) LHS = (sin θ - 2 sin³ θ) / (2 cos³ θ - cos θ)

= (sin θ(1 - 2 sin² θ)) / (cos θ(2 cos² θ - 1))

= (sin θ(sin² θ + cos² θ - 2 sin² θ)) / (cos θ(2 cos² θ - sin² θ - cos² θ))

= (tan θ(cos² θ - sin² θ)) / (cos² θ - sin² θ)

= tan θ = RHS.

(viii) LHS = (sin A + cosec A)² + (cos A + sec A)²

= sin² A + cosec² A + 2 + cos² A + sec² A + 2

= 4 + 1 + 1 + cot² A + 1 + tan² A

= 7 + tan² A + cot² A = RHS.

(ix) LHS = (cosec A - sin A)(sec A - cos A)

= (1/sin A - sin A)(1/cos A - cos A)

= ((1 - sin² A) / sin A)((1 - cos² A) / cos A)

= (cos² A / sin A)(sin² A / cos A) = sin A cos A

RHS = 1 / (tan A + cot A) = 1 / ((sin A / cos A) + (cos A / sin A))

= 1 / ((sin² A + cos² A) / (sin A cos A)) = (sin A cos A) / 1 = sin A cos A

∴ LHS = RHS.

(x) LHS = (1 + tan² A) / (1 + cot² A) = sec² A / cosec² A = tan² A = RHS.

((1 - tan A) / (1 - cot A))² = (((cos A - sin A) / cos A) / ((sin A - cos A) / sin A))²

= (sin² A / cos² A) = tan² A = RHS.

5.0Benefits of Class 10 Maths NCERT Solutions Ch 8 - Introduction to Trigonometry Exercise 8.4

  • NCERT solutions offer detailed, step-by-step explanations, making it easier for students to grasp concepts and learn the correct approach to solving trigonometric problems.
  • Exercise 8.4 focuses on trigonometric identities, helping students understand and apply fundamental trigonometric formulas in different mathematical problems.
  • By solving a variety of questions, students develop logical reasoning and problem-solving abilities, which are essential for higher-level mathematics and competitive exams.
  • Trigonometry is a crucial topic in exams like JEE, NDA, and other competitive exams. NCERT solutions ensure that students build a strong base for future studies.

NCERT Class 10 Maths Ch 8 Introduction to Trigonometry Other Exercises:

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

Frequently Asked Questions

Exercise 8.4 focuses on trigonometric identities, which are fundamental equations involving trigonometric ratios that hold true for all values of the variables.

Trigonometric identities are essential for: Simplifying trigonometric expressions. Solving complex trigonometric equations. Applications in coordinate geometry, physics, and calculus. Competitive exams like JEE, NTSE, and Olympiads.

Proof-based questions where you verify trigonometric identities. Expression simplification using trigonometric identities. Application-based problems involving fundamental trigonometric equations.

Yes! Questions from trigonometric identities frequently appear in Class 10 board exams and carry significant weightage. Understanding them well ensures scoring high in exams.

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