NCERT Solutions for Exercise 8.2 Class 10 Maths Chapter 8 - Introduction to Trigonometry

# NCERT Solutions for Exercise 8.2 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Edited By Ramraj Saini | Updated on Nov 27, 2023 08:56 AM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2

NCERT Solutions for Exercise 8.2 Class 10 Maths Chapter 8 Introduction to Trigonometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 8.2 deal trigonometric ratios and ratios of particular specific angles such as 0, 30, 45, 60 and 90 degrees. The ratios of the sides of a right triangle to their sharp angles are known as trigonometric ratios. The hypotenuse, which is the longest side of the right angle triangle, the perpendicular, which is the opposite side to the right angle triangle's angle, and the base, which is the next side to the right angle triangle's angle are the three sides of the right angle triangle.

NCERT solutions for exercise 8.2 Class 10 Maths chapter 8 Introduction to Trigonometry focuses on trigonometry and trigonometric ratios including sine, cosine, tangent, cosecant, secant and cotangent for specific angle. 10th class Maths exercise 8.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 8 exercise 8.2

Introduction to Trigonometry class 10 chapter 8 Exercise: 8.2

$(i) \sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}$

$\sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}$
As we know,
the value of $\sin 60^0 = \sqrt{3}/2 = \cos 30^0$ , $\sin 30^0 = 1/2=\cos 60^0$
$\\\Rightarrow \frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+ \frac{1}{2}.\frac{1}{2}\\$
$=\frac{3}{4}+\frac{1}{4}$
$=1$

$(ii)\: 2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}$

We know the value of

$\tan 45^0 = 1$ and

$\cos 30^0 = \sin 60^0 = \frac{\sqrt{3}}{2}$
According to question,

$=2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}$
$\\=2(1)^2+ (\frac{\sqrt{3}}{2})^2-(\frac{\sqrt{3}}{2})\\=2$

$(iii)\: \frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}$

$\frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}$
we know the value of

$\cos 45 = 1/\sqrt{2}$ , $\sec 30^0 = 2/\sqrt {3}$ and $cosec \:30 =2$ ,

After putting these values
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{1/\sqrt{3}}{(2+2\sqrt{3})/ \sqrt{3}}$
$\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}$
$\\=\frac{2\sqrt{6}-2\sqrt{18}}{-16}\\ =2\frac{\sqrt{6}-3\sqrt{3}}{-16} = \frac{3\sqrt{3}-\sqrt{6}}{8}$

$(iv)\: \frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$

$\frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$ ..................(i)
It is known that the values of the given trigonometric functions,
$\\\sin 30^0 = 1/2=cos 60^0\\ \tan 45^0 = 1=\cot 45^0\\ \sec 30^0 = 2/\sqrt{3}=cosec 60^0\\$
Put all these values in equation (i), we get;
$\\\Rightarrow \frac{1/2+1-2/\sqrt{3}}{2/\sqrt{3}+1/2+1}\\ \Rightarrow\frac{3/2-2/\sqrt{3}}{3/2+2/\sqrt{3}}\\ \Rightarrow \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}\\ \Rightarrow \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}\\ \Rightarrow \frac{43-24\sqrt{3}}{11}$

$(v)\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}$

$\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}$ .....................(i)
We know the values of-
$\\\cos 60^0 = 1/2= \sin 30^0\\ \sec 30^0 = 2/\sqrt{3}\\ \tan 45^0 = 1\\ \cos 30^0 = \sqrt{3}/2$
By substituting all these values in equation(i), we get;

$\\\Rightarrow \frac{5.(\frac{1}{2})^2+4.(\frac{2}{\sqrt{3}}) ^2-1}{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}\\ \Rightarrow \frac{5/4-1+16/3}{1}\\ \Rightarrow \frac{1/4+16/3}{1}\\ \Rightarrow\frac{67}{12}$

$(i)\, \frac{2\: \tan 30^{o}}{1+\tan ^{2}30^{o}}=$

$(A)\: \sin 60^{o}$ $(B)\: \cos 60^{o}$ $(C)\: \tan 60^{o}$ $(D)\: \sin 30^{o}$

Put the value of tan 30 in the given question-
$\frac{2\: \tan 30^{o}}{1+\tan ^{2}30^{o}}=\frac{2\times1/\sqrt{3}}{1+(1/\sqrt{3})^2}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\sqrt{3}/2 = \sin 60^0$

The correct option is (A)

$(ii)\: \frac{1-\tan^{2}45^{o}}{1+ \tan^{2}45^{o}}=$

$(A)\: \tan \: 90^{o}$ $(B)\: 1$ $(C) \: \sin 45^{o}$ $(D) \: 0$

The correct option is (D)
$\frac{1-\tan^{2}45^{o}}{1+ \tan^{2}45^{o}}=$
We know that $\tan 45 = 1$
So, $\frac{1-1}{1+1}=0$

$(iii)\sin \: 2A=2\: \sin A$ is true when $A$ =

$(A)0^{o}$ $(B)\: 30^{o}$ $(C)\: 45^{o}$ $(D)\: 60^{o}$

The correct option is (A)
$\sin \: 2A=2\: \sin A$
We know that $\sin 2A = 2\sin A \cos A$
So, $2\sin A \cos A = 2\sin A$
$\\\cos A = 1\\ A = 0^0$

$(iv)\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=$

$(A)\: \cos 60^{o}$ $(B)\: \sin 60^{o}$ $(C)\: \tan 60^{o}$ $(D)\: \sin 30^{o}$

Put the value of $tan 30^{o}={\1/\sqrt{3}}$

$\\=\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=\frac{2\times 1/\sqrt{3}}{1-(\frac{1}{\sqrt{3}})^2}\\ =\frac{2/\sqrt{3}}{1-1/3}\\ =\frac{2}{\sqrt{3}}\times \frac{3}{2}\\ =\sqrt{3}= \tan 60^0$

The correct option is (C)

Given that,
$\tan (A+B) = \sqrt{3} =\tan 60^0$
So, $\dpi{100} A + B = 60^o$ ..........(i)
$\tan (A-B) = 1/\sqrt{3} =\tan 30^0$
therefore, $\dpi{100} A - B = 30^o$ .......(ii)
By solving the equation (i) and (ii) we get;

$\dpi{100} A = 45^o$ and $\dpi{100} B = 15^o$

$(i) \sin (A + B) = \sin A + \sin B$
$(ii)$ The value of $\sin \theta$ increases as $\theta$ increases.
$(iii)$ The value of $\cos \theta$ increases as $\theta$ increases.
$(iv)\sin \theta =\cos \theta$ for all values of $\theta$ .
$(v) \cot A$ is not defined for $A=0^{o}$

(i) False,
Let A = B = $45^0$
Then, $\\\sin(45^0+45^0) = \sin 45^0+\sin 45^0\\ \sin 90^ = 1/\sqrt{2}+q/\sqrt{2}\\ 1 \neq \sqrt{2}$

(ii) True,
Take $\theta = 0^0,\ 30^0,\ 45^0$
whent
$\theta$ = 0 then zero(0),
$\theta$ = 30 then value of $\sin \theta$ is 1/2 = 0.5
$\theta$ = 45 then value of $\sin \theta$ is 0.707

(iii) False,
$\cos 0^0 = 1,\ \cos 30^0 = \sqrt{3}/2= 0.87,\ \cos 45^0 = 1\sqrt{2}= 0.707$

(iv) False,
Let $\theta$ = 0
$\\\sin 0^0 = \cos 0^0\\ 0 \neq 1$

(v) True,
$\cot 0^0 = \frac{\cos 0^0}{\sin 0^0}=\frac{1}{0}$ (not defined)

## More About NCERT Solutions for Class 10 Maths Exercise 8.2:

The link between the trigonometric ratios was also covered in the NCERT solutions for Class 10 Maths exercise 8.2. The Trigonometric ratios of particular specific angles were the focus of NCERT book exercise 8.2 Class 10 Maths. The Trigonometric ratios of some specific angles are the ratios of the sides of a right-angle triangle with regard to any of its acute angles. The trigonometric ratios of some specific angles are 0°, 30°, 45°, 60° and 90°. The following table lists the six trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°.

 Sin A 0 1 Cos A 1 0 Tan A 0 1 Not defined Cosec A Not defined 2 1 Sec A 1 2 Not defined Cot A Not defined 1 0

Usage of Pythagoras theorem is also covered in Exercise 8.2 Class 10 Maths. Students can use these Introduction to Trigonometry Class 10 notes to quick revision of the important concepts discussed in this chapter.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 8.2:

• NCERT solutions for Class 10 Maths exercise 8.2 can be used in astronomy to measure the distance to nearby stars as well as in satellite navigation systems.
• Exercise 8.2 Class 10 Maths, is also used in the calculation of lengths, areas, and relative angles between objects.
• NCERT solution for Class 10 Maths chapter 8 exercise 8.2 will help students attain a good score in term exams and help in further higher studies.

Also, see-

## Subject Wise NCERT Exemplar Solutions

1. What is the value of cos60° sin30° + sin60°cos30° ?

cos60° sin30° + sin60°cos30°

=1/4+3/4=4/4

=1

2. What is the value of cos 3 π?

The value of cos 3π is cos(π+2π)

=cos(π) =-1

3. What is the value of tan 45° ?

The value of tan 45° is 1

4. State true or false. The trigonometric function that is equal to the ratio of the side opposite a given angle to the hypotenuse is called sine.

Yes, the given assertion is correct. The sine is the ratio of the opposing side to the hypotenuse of a right-angle triangle.

Sin θ denotes the opposite side/hypotenuse

5. Say yes/no. Whether the value of cos 30° is equal to sin 60°

Yes, the value of cos 30° is equal to sin 60°.

6. What is the value of cosec 0° ?

The value of cosec 0° is not defined.

7. According to NCERT solutions for Class 10 Maths chapter 8 exercise 8.2 , What are the trigonometric ratios of specific angles?

The Trigonometric ratios of some specific angles are the ratios of the sides of a right-angle triangle with regard to any of its acute angles.

8. According to NCERT solutions for Class 10 Maths chapter 8 exercise 8.2 , what are the six trigonometric ratios ?

The six trigonometric ratios are Sine  , Cosine , Tangent , Cotangent , Cosecant   and Secant.

9. How many questions are there in the NCERT solutions for Class 10 Maths chapter 8 exercise 8.2 and what types of questions are there ?

NCERT solutions for Class 10 Maths chapter 8 exercise sides 8.2 consist of 4 questions and all the 4 questions are based on trigonometric ratios of specific angles.

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