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

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The exercise provides students with an evaluation of trigonometric ratios at standard angles ranging from 0° to 30° to 45° to 60° to 90°. The exercise develops mastery of trigonometric ratios behaviour during this angle range since trigonometry depends on this fundamental principle. Studying these problems allows us to become efficient at recalling and applying these values, which they will need for solving complex trigonometric problems in advanced mathematics courses.

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NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.2
Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.2
Topics Covered in Chapter 8, Introduction to Trigonometry: Exercise 8.2
NCERT Solutions of Class 10 Subject Wise
NCERT Exemplar Solutions of Class 10 Subject Wise

exercise 8.2

Theory application from trigonometric ratios comes into focus within this part of the NCERT Solutions for Class 10 Maths, which follows the NCERT book material. The students engage with problems where they determine the values of sine and cosine, and tangent functions while developing an understanding of how these trigonometric ratios connect to each other. The completion of Exercise 8.2 in the NCERT Books allows students to deepen their understanding of trigonometry and create essential analytical skills needed for higher mathematics education.

NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.2

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Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.2

Q 1: Evaluate the following :

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

Answer:

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

Q 1: Evaluate the following :

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

Answer:

We know the value of

$\tan 45^{0} = 1$ and

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

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

Q 1: Evaluate the following :

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

Answer:

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

$\cos 45 = 1 / \sqrt{2}$ , $\sec 30^{0} = 2 / \sqrt{3}$ and $c o s e c 30 = 2$ ,

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

Q 1: Evaluate the following :

$(i v) \frac{\sin 30^{o} + \tan 45^{o} - c o s e c 60^{o}}{\sec 30^{o} + \cos 60^{o} + \cot 45^{o}}$

Answer:

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

Q 1: Evaluate the following :

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

Answer:

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

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

Q 2: Choose the correct option and justify your choice :

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

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

Answer:

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

The correct option is (A)

Q 2: Choose the correct option and justify your choice :

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

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

Answer:

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

Q 2: Choose the correct option and justify your choice :

$(i i i) \sin 2 A = 2 \sin A$ is true when $A$ =

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

Answer:

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

Q 2: Choose the correct option and justify your choice :

$(i v) \frac{2 \tan 30^{o}}{1 - t a n^{2} 30^{o}} =$

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

Answer:

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

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

The correct option is (C)

Q 3: If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \frac{1}{\sqrt{3}};$ $0^{o} < A + B \leq 90^{o}; A > B,$ find $A a n d B .$

Answer:

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

$A = 45^{o}$ and $B = 15^{o}$

Q 4: State whether the following are true or false. Justify your answer.

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

Answer:

(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 $θ = 0^{0}, 30^{0}, 45^{0}$
whent
$θ$ = 0 then zero(0),
$θ$ = 30 then value of $\sin θ$ is 1/2 = 0.5
$θ$ = 45 then value of $\sin θ$ 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 $θ$ = 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)

Also Read-

Topics Covered in Chapter 8, Introduction to Trigonometry: Exercise 8.2

1. Trigonometric Ratios of Standard Angles: Evaluating sine, cosine, and tangent for 0°, 30°, 45°, 60°, and 90°.

2. Complementary Angles: Understanding the relationship between trigonometric ratios of complementary angles.

3. Application of Trigonometric Values: The practice of solving problems requires applying standard trigonometric values.

4. True or False Statements: The assessment deals with true or false statements about trigonometric ratios to confirm their validity while providing justification for the correct answers.

5. Multiple-Choice Questions: Students must choose appropriate trigonometric ratios from lists of options through calculations.

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Frequently Asked Questions (FAQs)

Q: What is the value of cos60° sin30° + sin60°cos30° ?

cos60° sin30° + sin60°cos30°

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

Q: What is the value of cos 3 π?

The value of cos 3π is cos(π+2π)

=cos(π) =-1

Q: What is the value of tan 45° ?

The value of tan 45° is 1

Q: 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

Q: 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°.

Q: What is the value of cosec 0° ?

The value of cosec 0° is not defined.

Q: 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.

Q: 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.

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

NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.2

Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.2

Topics Covered in Chapter 8, Introduction to Trigonometry: Exercise 8.2

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