NCERT Solutions of Class 10 Maths Chapter 8: Introduction to Trigonometry

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NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry are given in the article. Trigonometry studies relationships between side lengths and angles of triangles. Chapter 8 Introduction to Trigonometry solutions cover key concepts such including Trigonometric ratios of an acute angle of a right-angled triangle & their proofs. Values of the trigonometric ratios of 30⁰, 40⁰, and 60⁰ and relationships between Ratios.

Class 10 Maths Chapter 8 Introduction to Trigonometry is a part of Unit 5 Trigonometry. This unit holds a weightage of 12 Marks in Class 10 Maths Examination 2022-23. NCERT Solutions of Class 10 Maths Chapter 8 are based on the following topics:

Download: NCERT Solutions for Class 12 Mathametics Chapter 8 pdf

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry are provided below:

Also check: Introduction to Trigonometry

Important Topics: NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry

Important Topics of NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry are elaborated below:

Trigonometric Identities

There are 6 kinds of Trigonometric Idenities:

Sine
Cosine
Tangent
Secant
Cosecant
Cotangent

Example Prove the following identity using the trigonometric identities:

[(sin 3θ + cos 3θ)/(sin θ + cos θ)] + sin θ cos θ = 1

Solution: Let’s the following identity:

a³+b³ = (a+b)(a²-ab+b²)

Using Pythagoras Theorem,

L.H.S. = [(sin 3θ + cos 3θ)(sin θ + cos θ)] + sin θ cos θ

= [(sin θ + cos θ)(sin2θ - sin θ cos θ + cos2θ)(sin θ + cos θ) + sin θ cos θ

= (sin2θ - sin θ cos θ + cos2θ) + sin θ cos θ

= sin2θ + cos2θ = 1 = R.H.S.

Trigonometry Table

Trigonometry Table is a collection of values of trigonometric ratios for various standard angles including 0°, 30°, 45°, 60°, 90°, sometimes with other angles like 180°, 270°, and 360° included, in a tabular format.

Example: Calculate the exact value of sin15º using the trigonometric value for standard angles from a trigonometry table.

Solution: Using the trigonometric table, we know that sin45º = 1/√2, cos30º = (√3/2), cos45º = 1/√2, and sin30º = 1/2

sin15º = sin(45º - 30º) = sin45ºcos30º - cos45ºsin30º = (√2/2) • (√3/2) - (√2/2) • (1/2) = (√6 - √2)/4 = (√3 - 1)/2√2

Therefore, tThe value of sin15º = (√3 - 1)/2√2

Trigonometry Ratios

Trigonometric ratios are the “ratios of length of sides of a triangle”.

Trigonometric ratios relate the ratio of sides of a right triangle to the respective angle. Basic trigonometric ratios are sin, cos, and tan. Other important trigonometric ratios, cosec, sec, and cot, can be derived using the sin, cos, and tan respectively.

Sum, Difference, Product Trigonometric Ratios Identities

Sum, Difference, Product Trigonometric Ratios include the following formulas:

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

NCERT Solutions For Class 10 Maths Chapter 8 Exercises

The detailed solutions for all the NCERT Solutions for Introduction to Trigonometry under different exercises are as follows:

Also check:

Important Chapter Related Topics
Introduction to Trigonometry Important Questions	Introduction to Trigonometry Formula	Introduction to trigonometry Revision Notes
Trigonometry Table	Cos 0	Value of cos 60
Sin 30 Degrees	sec 90	30 60 90 formula
cosec cot formula	sec 60	Difference between Trigonometry and Geometry

Also check:

Maths Study Guides
Math MCQs	Trigonometry	Mensuration
Geometry	Algebra	Difference Between in Maths
Maths Study Material	Math Formula	Class 10 Maths Study Notes

CBSE X Related Questions

1.
In the figure given above, $\triangle ABC \sim \triangle XYZ$, then find the values of $x$ and $y$.
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2.
Assertion (A) : H.C.F. $(36 m^{2}, 18 m) = 18 m$, where $m$ is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
3.
If $\alpha, \beta$ are the zeroes of the polynomial $p(x) = x^2 - 3x - 1$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
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4.
If the pair of linear equations : $ a_1x + b_1y + c_1 = 0 $ and $ a_2x + b_2y + c_2 = 0 $ is consistent and dependent, then
- $ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $
- $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $
- $ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $
- $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $
View Solution
5.
Two water taps together can fill a tank in $8\frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
View Solution
6.
Prove that :
$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$.
View Solution

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NCERT Solutions of Class 10 Maths Chapter 8: Introduction to Trigonometry

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry