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

NCERT Solutions for Class 10 Maths Exercise 8.2 of Chapter 8 Introduction to Trigonometry are mandatory for the preparation of CBSE board exams of 10 class .You can clear all your doubts about Introduction to Trigonometry by the study of these NCERT Solutions of exercise 8.1 Chapter 8.NCERT Solutions for Class 10 Maths Exercise 8.2 of Chapter 8 Introduction to Trigonometry are created by an expert of maths who has huge experience of maths teaching from class 6 to 12 classes.We have been teaching academic classes from class 1 to 12 in our private coaching centre for last 25 years in New Delhi,therefore you are free to ask us your problems in study we will definitely solve your all obstacles on your ways through our huge experience in the field of education.

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**NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry**

**Exercise 8.1- Introduction to Trigonometry**

**Exercise 8.2 -Introduction to Trigonometry**

**Exercise 8.3-Introduction to Trigonometry**

**Exercise 8.4-Introduction to Trigonometry**

**NCERT solutions of Important Questions-Introduction to Trigonometry**

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

**Q1.Evaluate the following**

**(i) sin 60°cos 30°+ sin 30°cos 60°**

**(ii) 2 tan²45°+ cos²30°- sin² 60°**

Ans. (i) sin 60°cos 30°+ sin 30°cos 60°

Putting the following values in the given expression

sin 60°=√3/2,cos 30°=√3/2, sin 30°=1/2,cos 60°= 1/2

(√3/2) ×(√3/2) + (1/2)×(1/2)

=(3/4) + (1/4)

=4/4 = 1

(ii) 2 tan²45°+ cos²30°- sin² 60°

Putting the values of tan45°=1, cos30°=√3/2, sin 60°=√3/2

2×1² +(√3/2)² -(√3/2)²

=2 + 3/4 -3/4 = 2

(iii) cos 45°/(sec30°+ cosec 30°)

Putting the values of cos 45° = 1/√2, sec 30° =2/√3, cosec 30° =2 in the given expression

(iv) The given expression is

(sin 30° + tan 45°-cosec 60°)/(sec 30° + cos 60°+cot 45°)

Putting the values of sin 30°=1/2, tan 45°=1,cosec 60°=2/√3,sec 30°=2/√3 , cos 60°=1/2,cot 45°=1 in the given expression

Rationalizing the denominator

Putting the values of cos 60° = 1/2, sec 30° = 2/√3 , tan 45° =1, sec 30° = 2/√3, cos 30° = √3/2

**Q2. Choose the correct options and justify your choice.**

**(i) 2 tan 30°/(1+ tan²30°)=**

**(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°**

**(ii) (1 – tan²45°)/ (1 – tan²45°)=**

**(A) tan 90° (B) 1 (C) sin 45° (D) 0**

Ans (A) is correct

Substitute tan 30° in the given equation

tan 30° = 1/√3

2 tan 30°**/** 1 + tan² 30° = 2(1/√3)**/**1 + (1/√3)²

= (2/√3)**/**(1 + 1/3) = (2/√3)**/**(4/3)

= 6**/**4√3 = √3**/**2 = sin 60°

The obtained solution is equivalent to the trigonometric ratio sin 60°

(ii) D is correct

Substitute tan 45° in the given equation

tan 45° = 1

1 – tan² 45°**/**1 + tan² 45° = (1 – 1²)**/**(1 + 1²)

= 0**/**2 = 0

The solution of the above equation is 0

(iii) (A) is correct

To find the value of A, substitute the degree given in the options one by one sin 2A = 2 sin A is true when A = 0°

As sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

(iv) (C) is correct

Substitite tan 30° in the given equation

tan 30° = 1/√3

2tan30°**/**1 – tan²30° = 2(1**/**√3)**/**1 – (1**/**√3)²

= (2/√3)**/**(1 – 1/3) = (2/√3)**/**(2/3) = √3 = tan 60°

The value of the given equation is equivalent to tan 60°

**Q3. If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, find A and B**

Ans. tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A +B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A = 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

**Q4. State whether the following are true or false. Justify your answer**

**(i) sin (A + B) = sin A + sin B**

**(ii) The value of sin θ increases as θ increases**

**(iii) The value of cos θ increases as θ increases**

**(iv) sin θ = cos θ for all values of θ**

**(v) cot A is not defined for A = 0°**

Ans. False

Justification:

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we get

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 6o°

= 1**/**2 + √3**/**2 = 1 + √3**/**2

Since the values obtained are not equal, the solution is false

(ii) True

Justification:

According to the values obtained as per the unit circle, the values of sin are:

sin 0° = 0

sin 30° = 1**/**2

sin 45° = 1**/**√2

sin 60° = √3**/**2

sin 90° = 1

Thus the value of sin θ increases as θ increases. Hence, the statement is true

(ii) False

Justification:

According to the values obtained as per the unit circle, the values of cos are:

cos 0° = 1

cos 30° = √3**/**2

cos 45° = 1**/**√2

cos 60° = 1**/**2

cos 90° = 0

Thus the value of cos θ increases as θ increases. Hence, the statement is false

(iv) False

sin θ = cos θ, when a right triangle has 2 angles of (π/4). Therefore, the above statement is false

(v) True

Since cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A**/**sin A

Now substitute A = 0°

cot 0° = cos 0°**/**sin 0° = 1**/**0 = undefined

Hence, it is true

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |