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

NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 will help students polish their problem-solving skills on the topic of trigonometric ratios of complementary angles. The exercise will focus on the important relation between trigonometric ratios and their complementary angles. The idea is a helpful tool in solving some of the intricate trigonometric issues, as well as some clever practical uses. So, let's start discovering this important subject of trigonometry, which will eventually be a strong foundation for further math studies.

1.0Download NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 : Free PDF 

2.0Introduction to Trigonometric Ratios of Complementary Angles:

Before directly moving to the key concepts and formulas related to complementary angles, let’s understand the meaning of these angles and their importance in trigonometry. We know from geometry that two angles are said to be complementary when the sum of both those angles is equal to 90° . Similarly, in the context of trigonometry, every trigonometric ratio is equal to certain angles; these angles, when written in the form of their complementary angles, change their value. In simple words, in a right-angled triangle, the sum of the non-right angles adds up to 90° and, hence, is known as the complementary angles of trigonometric ratios. It is a foundational concept in Exercise 8.3 for the conversion of different trigonometric ratios into one another. 

3.0Key Concepts of Exercise 8.3: Overview 

Exercise 8.3 primarily focuses on the conversion techniques of trigonometric ratios via the concepts of their complementary angles. Let’s dive into some key concepts of this exercise: 

Trigonometric Ratios for an Angle A:

Understanding the trigonometric ratios for an angle A in a right-angled triangle ABC, with base AB, will help you properly grasp the concept of complementary angles of these ratios. Below are the six basic ratios of trigonometry: 

1) $\sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

2) $\cos A=\frac{\text{Base}}{\text{Hypotenuse}}$

3) $\tan A=\frac{\text{Perpendicular}}{\text{Base}}$

4) $\mathrm{cosec}A=\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

5) $\sec A=\frac{\text{Hypotenuse}}{\text{Base}}$

6) $\cot A=\frac{\text{Base}}{\text{Perpendicular}}$

Trigonometric Ratios for the Complementary Angle: 

From the above definitions of complementary angles, we know that complementary angles are the sum of two non-right angles of a right-angled triangle. In the previous example, the complementary angle can be written as A + B = 90 or B = 90 – A. Hence, the trigonometric ratios will change as: 

1) $\sin (90-A)=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

2) $\cos (90-A)=\frac{\text{Base}}{\text{Hypotenuse}}$

3) $\tan (90-A)=\frac{\text{Perpendicular}}{\text{Base}}$

4) $\mathrm{cosec}(90-A)=\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

5) $\sec (90-A)=\frac{\text{Hypotenuse}}{\text{Base}}$

6) $\cot (90-A)=\frac{\text{Base}}{\text{Perpendicular}}$

Conversion of Trigonometric Ratios: 

In Exercise 8.3, every question is based on the conversion of trigonometric ratios into one another. These relationships indicate that for any angle A, the trigonometric ratios of the complementary angle 90° – A are simply the reciprocal of the corresponding ratios for A. Remember the following conversion formulas to solve questions in this exercise: 

1) Sin(90-A)=CosA 

2) Cos(90-A)=SinA 

3) tan(90-A)=CotA 

4) Cosec(90-A)=SecA 

5) Sec(90-A)=CosecA 

6) Cot(90-A)=tanA

Undefined Trigonometric Ratios: 

Note that not all the values of complementary angles have valid values. Certain trigonometric ratios are undefined for particular angles. Some examples of such values are: 

• tan90° and cot90° are undefined, as the value of the angle will become 0, which is an undefined value, as per the values of specific angles. 
• Similarly, sec⁡90° and cosec⁡0° are undefined due to division by zero.

To clarify your concept and excel in this exercise, begin practising using the NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 from the correct source.

4.0NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3: Detailed Solutions

1. Evaluate:

(i) sin 18° / cos 72°

(ii) tan 26° / cot 64°

(iii) cos 48° - sin 42°

(iv) cosec 31° - sec 59°

Solution:

(i) sin 18° / cos 72° = sin 18° / cos (90° - 18°) = sin 18° / sin 18° = 1

{∵ cos (90° - θ) = sin θ}

(ii) tan 26° / cot 64° = tan (90° - 64°) / cot 64° = cot 64° / cot 64° = 1

(iii) cos 48° - sin 42° = cos (90° - 42°) - sin 42° = sin 42° - sin 42° = 0

(iv) cosec 31° - sec 59° = cosec (90° - 59°) - sec 59° = sec 59° - sec 59° = 0

2. Show that

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

(ii) cos 38° cos 52° - sin 38° sin 52° = 0.

Solution:

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

= tan 48° × tan 23° × tan (90° - 48°) × tan (90° - 23°)

= tan 48° × tan 23° × cot 48° × cot 23°

= tan 48° × tan 23° × (1 / tan 48°) × (1 / tan 23°) = 1

∴ LHS = RHS.

(ii) LHS = cos 38° cos 52° - sin 38° sin 52°

= cos (90° - 52°) cos 52° - sin (90° - 52°) sin 52°

= sin 52° cos 52° - cos 52° sin 52° = 0

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

Solution:

tan 2A = cot (A - 18°)

=> cot (90° - 2A) = cot (A - 18°)

=> 90° - 2A = A - 18°

=> 108° = 3A

A = 36°

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

Solution:

tan A = cot B

tan A = tan (90° - B)

∴ A = 90° - B

A + B = 90°

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

Solution:

sec 4A = cosec (A - 20°)

=> cosec (90° - 4A) = cosec (A - 20°)

{∵ cosec (90° - θ) = sec θ}

=> 90° - 4A = A - 20°

=> 5A = 110°

=> A = 22°

6. If A, B and C are interior angles of a triangle ABC, then show that sin ((B + C) / 2) = cos (A / 2).

Solution:

A + B + C = 180°

=> B + C = 180° - A

=> (B + C) / 2 = (180° - A) / 2

=> (B + C) / 2 = 90° - (A / 2)

=> sin ((B + C) / 2) = sin (90° - (A / 2)) = cos (A / 2)

{∵ sin (90° - θ) = cos θ}

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

sin 67° + cos 75°

= sin (90° - 23°) + cos (90° - 15°)

= cos 23° + sin 15°

5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3

1. This helps students learn the correct approach to solving problems and understand why each step is taken.
2. Regular practice boosts analytical thinking and improves accuracy in exams.
3. Solving NCERT solutions ensures students are well-prepared and familiar with the types of questions likely to appear.