Introduction to Trigonometry

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.

Table of Content

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.

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.

(b) \(\frac{2}{3}\)

(c) \(\frac{1}{3}\)

(d) \(\frac{4}{3}\)

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

**Explanation:**cos

⇒ (cos

⇒ (cos

⇒ cos

⇒ 1- sin

⇒ 1 - 2 sin

(b)\(\frac{2}{3}\)

(c) \(\frac{3}{2}\)

(d)\(\frac{5}{2}\)

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

**Explanation:**Given : tan A =\(\frac{1}{\sqrt{5}}\)

⇒ \(\Rightarrow tan^2{A}=\frac{1}{5}...{i}\)

⇒ \(\Rightarrow \frac{1}{cot^2A}=\frac{1}{5}\begin{bmatrix} ∵ tanA = \frac{1}{cot A}\end{bmatrix}\)

⇒ cot

Now,\(\frac{cosec^2A-sec^2A}{cosec^2A+sec^2A}=\frac{1+cot^2A-(1+tan^2A)}{1+cot^2A+1+tan^2A}\)

[∵ cosec

\(=\frac{cot^2A-tan^2A}{2+cot^2A+tan^2A}=\frac{5-{\large\frac{1}{5}}}{2+5+{\large\frac{1}{5}}}[ From (i) and (ii)]\)

\(=\frac{\large\frac{25-1}{5}}{\large\frac{35+1}{5}}=\frac{24}{36}=\frac{2}{3}\)

\(\frac{cosec^2A-sec^2}{cosec^2A+sec^2}=\frac{2}{3}\)

\(\begin{pmatrix} \frac{1+tan^2A}{1+cot^2A} \end{pmatrix}=\begin{pmatrix} \frac{1-tanA}{1cotA} \end{pmatrix}^2\)

**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),

\(\begin{pmatrix} \frac{1+tan^2A}{1+cot^2A}\end{pmatrix}=\begin{pmatrix} \frac{1-tanA}{1-cotA}\end{pmatrix}^2\)

\(=tan^2A\)

(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 \(\sqrt{a^2+b^2-c^2}= (a sin θ + b cos θ)\)

\(\Rightarrow (a sin θ + b cos θ) =\sqrt{a^2+b^2-c^2}\)

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

**Explanation:**L.H.S. = (tan A + cosec B)

= tan

= tan

= (cosec

= 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.

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.

**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

**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).

**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.

**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)

**Ans:** The Pythagorean Trigonometric Identities are:

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