Class 10 Mathematics Notes Chapter 8 (Chapter 8) – Examplar Problems (English) Book

Examplar Problems (English)
Alright class, let's focus on Chapter 8, 'Introduction to Trigonometry'. This chapter lays the foundation for trigonometry, which is frequently tested in various government examinations. Mastering these concepts is essential. We'll cover the core ideas from your NCERT Exemplar and textbook, keeping the competitive exam perspective in mind.

Chapter 8: Introduction to Trigonometry - Detailed Notes

1. Introduction

Trigonometry deals with the relationship between the angles and sides of triangles, specifically right-angled triangles in this introductory chapter.

2. Trigonometric Ratios

Consider a right-angled triangle ABC, right-angled at B. Let ∠ACB = θ (theta) be the angle we are considering.

  • Hypotenuse (H): The side opposite the right angle (AC).
  • Perpendicular (P) or Opposite Side: The side opposite to the angle θ (AB).
  • Base (B) or Adjacent Side: The side adjacent to the angle θ (BC).

(Remember: Perpendicular and Base depend on which angle (other than 90°) you are considering.)

The six trigonometric ratios are defined as follows:

  • sine θ (sin θ): Ratio of the Perpendicular to the Hypotenuse.
    sin θ = P/H = AB/AC
  • cosine θ (cos θ): Ratio of the Base to the Hypotenuse.
    cos θ = B/H = BC/AC
  • tangent θ (tan θ): Ratio of the Perpendicular to the Base.
    tan θ = P/B = AB/BC
  • cosecant θ (cosec θ): Ratio of the Hypotenuse to the Perpendicular. (Reciprocal of sin θ)
    cosec θ = H/P = AC/AB = 1/sin θ
  • secant θ (sec θ): Ratio of the Hypotenuse to the Base. (Reciprocal of cos θ)
    sec θ = H/B = AC/BC = 1/cos θ
  • cotangent θ (cot θ): Ratio of the Base to the Perpendicular. (Reciprocal of tan θ)
    cot θ = B/P = BC/AB = 1/tan θ

Mnemonic: A popular way to remember the primary ratios (sin, cos, tan) is "Pandit Badri Prasad / Har Har Bole" or "Some People Have / Curly Brown Hair / Turned Permanently Black".

S = P/H
C = B/H
T = P/B

Important Relationships:

  • tan θ = sin θ / cos θ
  • cot θ = cos θ / sin θ
  • cosec θ = 1 / sin θ => sin θ * cosec θ = 1
  • sec θ = 1 / cos θ => cos θ * sec θ = 1
  • cot θ = 1 / tan θ => tan θ * cot θ = 1

Key Points:

  • Trigonometric ratios are unitless numbers.
  • The value of sin θ and cos θ is always between -1 and 1 (inclusive). For angles between 0° and 90°, the value is between 0 and 1.
  • The value of sec θ and cosec θ is always ≤ -1 or ≥ 1. For angles between 0° and 90° (excluding 0° for cosec and 90° for sec), the value is ≥ 1.
  • tan θ and cot θ can take any real value.

3. Trigonometric Ratios of Specific Angles (0°, 30°, 45°, 60°, 90°)

Memorizing this table is crucial for quick calculations in competitive exams.

Angle (θ) 30° 45° 60° 90°
sin θ 0 1/2 1/√2 √3/2 1
cos θ 1 √3/2 1/√2 1/2 0
tan θ 0 1/√3 1 √3 Undefined
cosec θ Undefined 2 √2 2/√3 1
sec θ 1 2/√3 √2 2 Undefined
cot θ Undefined √3 1 1/√3 0
  • Tips for remembering:
    • Write numbers 0, 1, 2, 3, 4.
    • Divide by 4: 0/4, 1/4, 2/4, 3/4, 4/4.
    • Take square root: √0, √(1/4), √(1/2), √(3/4), √1. This gives sin values: 0, 1/2, 1/√2, √3/2, 1.
    • cos values are sin values in reverse order.
    • tan = sin/cos.
    • cosec, sec, cot are reciprocals of sin, cos, tan respectively.

4. Trigonometric Ratios of Complementary Angles

Two angles are complementary if their sum is 90°. In a right triangle ABC (right-angled at B), angles A and C are complementary (A + C = 90°).

The following relationships hold true:

  • sin (90° - θ) = cos θ
  • cos (90° - θ) = sin θ
  • tan (90° - θ) = cot θ
  • cot (90° - θ) = tan θ
  • sec (90° - θ) = cosec θ
  • cosec (90° - θ) = sec θ

Applications: These are used to simplify expressions like sin 18° / cos 72° or tan 26° cot 64°.
Example: sin 18° / cos 72° = sin 18° / cos (90° - 18°) = sin 18° / sin 18° = 1.

5. Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle(s) involved for which the ratios are defined.

Fundamental Identities:

  1. sin² θ + cos² θ = 1
    • This implies: sin² θ = 1 - cos² θ and cos² θ = 1 - sin² θ
  2. 1 + tan² θ = sec² θ (Derived by dividing the first identity by cos² θ)
    • This implies: sec² θ - tan² θ = 1 and tan² θ = sec² θ - 1
  3. 1 + cot² θ = cosec² θ (Derived by dividing the first identity by sin² θ)
    • This implies: cosec² θ - cot² θ = 1 and cot² θ = cosec² θ - 1

(These identities hold for 0° ≤ θ ≤ 90°, considering defined values)

Applications:

  • Simplifying trigonometric expressions.
  • Proving other trigonometric relationships.
  • Solving equations involving trigonometric ratios.

Example Problem Type (Government Exams):
If sin θ + cos θ = √2 cos θ, find cot θ.
Solution:
Divide by sin θ: 1 + cot θ = √2 cot θ
1 = √2 cot θ - cot θ
1 = cot θ (√2 - 1)
cot θ = 1 / (√2 - 1)
Rationalize: cot θ = (√2 + 1) / ((√2 - 1)(√2 + 1)) = (√2 + 1) / (2 - 1) = √2 + 1

Key Strategy for Proving Identities:

  • Start with the more complex side (LHS or RHS).
  • Express everything in terms of sin and cos.
  • Use algebraic manipulations (factorization, common denominators, etc.).
  • Apply the fundamental identities.
  • Sometimes, simplifying both LHS and RHS separately to a common expression helps.

Multiple Choice Questions (MCQs) for Practice

Here are 10 MCQs based on Chapter 8 concepts, relevant for government exam preparation:

1. In a right-angled triangle ABC, right-angled at B, if tan A = 1/√3, then the value of sin A is:
(a) √3/2
(b) 1/2
(c) 1/√2
(d) 1

2. The value of cos 0° * cos 30° * cos 45° * cos 60° * cos 90° is:
(a) 1
(b) √3/16
(c) 0
(d) 1/8

3. If sin θ = cos θ for an acute angle θ, then the value of 2 tan θ + cos² θ is:
(a) 1
(b) 2
(c) 3/2
(d) 5/2

4. The value of tan 10° tan 15° tan 75° tan 80° is:
(a) 0
(b) 1
(c) -1
(d) √3

5. If sin A = 3/5, then the value of 9 cot² A + 9 is:
(a) 9
(b) 25
(c) 16
(d) 5

6. (sec A + tan A)(1 – sin A) is equal to:
(a) sec A
(b) sin A
(c) cosec A
(d) cos A

7. If cos (α + β) = 0, then sin (α + β - 90°) can be reduced to:
(a) cos β
(b) cos 2β
(c) sin α
(d) -1

8. The value of (sin 30° + cos 30°) – (sin 60° + cos 60°) is:
(a) -1
(b) 0
(c) 1
(d) 2

9. If √3 tan θ = 1, then the value of sin² θ – cos² θ is:
(a) 1/2
(b) -1/2
(c) √3/2
(d) 2/√3

10. Given that sin α = 1/2 and cos β = 1/2, then the value of (α + β) is:
(a) 0°
(b) 30°
(c) 60°
(d) 90°


Answer Key:

  1. (b)
  2. (c)
  3. (d)
  4. (b)
  5. (b)
  6. (d)
  7. (d)
  8. (b)
  9. (b)
  10. (d)

Study these notes thoroughly, practice problems from your Exemplar book and previous year question papers of relevant government exams. Pay close attention to the identities and standard angle values. Good luck!

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