Introduction to Trigonometry

# Class 10 Math Chapter 8 Introduction To Trigonometry Important Questions  Here are some important questions for Class 10 Mathematics Chapter 8, Introduction to Trigonometry, carefully selected to help students prepare for the CBSE Class 10 Mathematics Examination in 2023-24. These questions cover various types of problems and are designed to assist students in understanding Trigonometry better. By practicing these diverse question types, students can clarify any doubts they may have and improve their problem-solving skills, leading to better performance in the chapter on Introduction to Trigonometry.

## Introduction

In Chapter 8 of Class 10 Mathematics, Introduction To Trigonometry, In this chapter, we will explore Trigonometric Ratios of acute angles in right-angled triangles. We will establish the existence and well-defined nature of these ratios. Additionally, we will understand why certain ratios are defined at 0° and 90°. We will also determine the values of trigonometric ratios for important angles like 30°, 45°, and 60°, and examine the relationships between these ratios.

### What is Trigonometry?

Trigonometry is a branch of mathematics that deals with the study of angles, triangles, and the relationships between the angles and sides of triangles. It is widely used in various fields, including physics, engineering, astronomy, and navigation. ### Class 10 Introduction To Trigonometry Important Questions and Answers

#### (a) 0(b) $$\frac{2}{3}$$(c) $$\frac{1}{3}$$(d) $$\frac{4}{3}$$

Ans. (b)$$\frac{2}{3}$$

Explanation:
cos4 θ - sin4 θ = $$\frac{2}{3}$$
⇒  (cos2 θ)2 - (sin2 θ)2 =$$\frac{2}{3}$$
⇒ (cos2 θ + sin2 θ)(cos2 θ - sin2 θ) =$$\frac{2}{3}$$
⇒     cos2 θ - sin2 θ =$$\frac{2}{3}$$        (∵ sin2 θ + cos2 θ = 1)
⇒ 1- sin2 θ - sin2 θ =$$\frac{2}{3}$$                        (∵ cos2 θ = 1-  sin2 θ)
⇒ 1 - 2 sin2 θ =$$\frac{2}{3}$$

#### (a)$$\frac{5}{3}$$(b)$$\frac{2}{3}$$(c) $$\frac{3}{2}$$(d)$$\frac{5}{2}$$

Ans. (b)$$\frac{2}{3}$$

Explanation:
Given : tan A  =$$\frac{1}{\sqrt{5}}$$
⇒ $$\Rightarrow tan^2{A}=\frac{1}{5}...{i}$$
⇒ $$\Rightarrow \frac{1}{cot^2A}=\frac{1}{5}\begin{bmatrix} ∵ tanA = \frac{1}{cot A}\end{bmatrix}$$
⇒ cot2 A   = 5 …(ii)
Now,$$\frac{cosec^2A-sec^2A}{cosec^2A+sec^2A}=\frac{1+cot^2A-(1+tan^2A)}{1+cot^2A+1+tan^2A}$$
[∵ cosec2 A = 1 + cot2 A and sec2 A = 1 + tan2 A]
$$=\frac{cot^2A-tan^2A}{2+cot^2A+tan^2A}=\frac{5-{\large\frac{1}{5}}}{2+5+{\large\frac{1}{5}}}[ From (i) and (ii)]$$
$$=\frac{\large\frac{25-1}{5}}{\large\frac{35+1}{5}}=\frac{24}{36}=\frac{2}{3}$$
$$\frac{cosec^2A-sec^2}{cosec^2A+sec^2}=\frac{2}{3}$$

#### Q3. Prove that $$\begin{pmatrix} \frac{1+tan^2A}{1+cot^2A} \end{pmatrix}=\begin{pmatrix} \frac{1-tanA}{1cotA} \end{pmatrix}^2$$

Ans. Proved

Explanation:
Consider,
$$\begin{pmatrix} \frac{1+tan^2A}{1+cot^2A} \end{pmatrix}=\frac{1+\frac{sin^2}{cos^2A}}{1+\frac{cos^2A}{sin^2A}}$$
$$=\begin{pmatrix} \frac{sin^2A+cos^2A}{sin^2A+cos^2A}\end{pmatrix}$$
$$\frac{sin^2A}{cos^2A}$$
$$=Tan^2A....(i)$$
$$\begin{pmatrix} \frac{1-tanA}{1-cotA} \end{pmatrix}^2=\begin{pmatrix} \frac{1-\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}} \end{pmatrix}$$
$$=\begin{bmatrix} \begin{pmatrix} \frac{cosA-sinA}{sinA-cosA}\end{pmatrix}\begin{pmatrix} \frac{sinA}{cosA}\end{pmatrix}\end{bmatrix}$$
$$\begin{bmatrix} \begin{pmatrix} -\frac{sinA-cosA}{sinA-cosA}\end{pmatrix}(tan A)\end{bmatrix}$$
$$=tan^2A$$......(ii)
From (i) and (ii),
$$\begin{pmatrix} \frac{1+tan^2A}{1+cot^2A}\end{pmatrix}=\begin{pmatrix} \frac{1-tanA}{1-cotA}\end{pmatrix}^2$$
$$=tan^2A$$

#### Q4. If a cos θ – b sin θ = c, prove that(a sin θ + b cos θ) =$$\sqrt{a^2+b^2-c^2}$$

Ans.  Proved

Explanation:
Given, a cos θ – b sin θ = c ...(i)
$$Now, a +b – c = a + b– (a cos θ – b sin θ) [From (i)]$$
= a + b – (acos θ + bsin θ – 2ab sin θ cos θ)
= a + b – [a(1 – sin θ) + b(1 – cos θ) – 2ab sin θ cos θ]
= a + b – [a – asin θ + b – b cos θ – 2ab sin θ cos θ]
= a + b – a+ asin θ – b + bcos θ + 2ab sin θ cos θ]
= asinθ + bcos θ + 2ab sin θ cos θ
=$$(a sin θ + b cos θ)^2$$
\hus, a + b – c = (a sin θ + b cos θ)^2
or $$\sqrt{a^2+b^2-c^2}= (a sin θ + b cos θ)$$
$$\Rightarrow (a sin θ + b cos θ) =\sqrt{a^2+b^2-c^2}$$

#### Q5. (tan A + cosec B) – (cot B – sec A) = 2 tan A cot B (cosec A + sec B).

Ans. L.H.S. = R.H.S.

Explanation:
L.H.S. = (tan A + cosec B)2 – (cot B – sec A)^2
= tan2 A + cosec2 B + 2 tan A.cosec B – (cot2 B + sec2 A – 2 cot B.sec A)
= tan2 A + cosec2 B + 2 tan A.cosec B – cot2 B – sec2 A + 2 cot B sec A
= (cosec2 B – cot2 B) – (sec2 A – tan2 A) + 2 tan A.cosec B + 2 cot B sec A
= 1 – 1 + 2 tan A.cosec B + 2 cot B.sec A
= 2 tan A.cosec B + 2 cot B.sec A
$$=2\begin{pmatrix} \frac {sinA}{cosA}×\frac{1}{sinB}+\frac{cosB}{sinB}×\frac{1}{cosA}\end{pmatrix}$$
$$=2\begin{pmatrix} \frac{sinA+cosB}{cosA.sinB}\end{pmatrix}...(i)$$
R.H.S. = 2 tan A.cot B.(cosec A + sec B)
= 2 tan A.cot B. cosec A + 2 tan A. cot B.sec B
$$=2\frac{sinA}{cosA}.\frac{cosB}{sinA}.\frac{1}{sinA}+\frac{2sinA}{cosA}.\frac{cosB}{sinB}.\frac{1}{cosB}$$
$$=2\begin{pmatrix} \frac{cosB}{cosA.sinB}+\frac{sinA}{cosA.sinB}\end{pmatrix}$$
$$=2 \begin{pmatrix} \frac{cosB+sinA}{cosA.sinB}....(ii)\end{pmatrix}$$
From (i) and (ii),
L.H.S. = R.H.S.  #### CBSE Class 10 Maths Chapter wise Important Questions

Chapter No. Chapter Name
Chapter 1 Real Number
Chapter 2 Polynomials
Chapter 3 Pair of Linear Equations in Two Variables
Chapter 5 Arithmetic Progressions
Chapter 6 Triangles
Chapter 7 Coordinate Geometry
Chapter 8 Introduction to Trigonometry
Chapter 9 Some Applications of Trigonometry
Chapter 10 Circles
Chapter 11 Areas Related to Circle
Chapter 12 Surface Areas and Volumes
Chapter 13 Statistics
Chapter 14 Probability

#### Conclusion

If you want to improve your understanding of the Introduction to Trigonometry chapter, you should check out oswal.io. This website provides a variety of practice questions designed to help you learn better. By working on these questions, you can strengthen your knowledge of trigonometry and become better at solving problems related to this important math topic. It's a great way to practice and get more comfortable with trigonometry.

#### Q1:  What are the Trigonometric Ratios?

Ans: Trigonometric ratios are ratios of the sides of a right-angled triangle. The three sides of the right triangle are:

• Hypotenuse (the longest side)
• Perpendicular (opposite side to the angle)
• Base (Adjacent side to the angle

The three main trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).They are defined as follows:

• Sine (sin) = Opposite side / Hypotenuse
• Trigonometric ratios are ratios of the sides
• Trigonometric ratios are ratios of the sides

#### Q2: How are Trigonometric Ratios related to 0° and 90°?

Ans: At 0°, the values of sine and tangent are 0, while the value of cosine is 1. At 90°, the value of sine is 1, the value of cosine is 0, and the value of tangent is undefined (approaches infinity).

#### Q3: What is the range of values for the sine, cosine, and tangent ratios?

Ans: The values of sine and cosine ratios lie between -1 and 1, inclusive. However, the tangent ratio can take any real value, including positive infinity and negative infinity, except when the angle is a multiple of 90°, where it becomes undefined.

#### Q4: What are the reciprocal trigonometric ratios?

Ans: The reciprocal trigonometric ratios are secant (sec), cosecant (cosec), and cotangent (cot). They are defined as follows:

• sec(θ) = 1 / cos(θ)
• cosec(θ) = 1 / sin(θ)
• cot(θ) = 1 / tan(θ)

#### Q5: What are the Pythagorean Trigonometric Identities?

Ans: The Pythagorean Trigonometric Identities are:

• sin²θ + cos²θ = 1
• tan²θ + 1 = sec²θ
• 1 + cot²θ = cosec²θ
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