**NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry**

NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry are presented here by Future Study Pointed free of cost. These NCERT Solutions of Exercise 8.3-Introduction to Trigonometry will boost your preparation of the forthcoming exams and in doing your homework and class assignments.You can study here all the exercises of the chapter 8-Introduction to Trigonometry.All questions are solved by an expert of maths as per the CBSE norms prescribed for class 10 standard.

The NCERT Solutions for class 10 maths are the most important study material for class 10 in clearing their doubts on maths. The marks in mathematics play an important role in increasing your aggregate percentage in the result of the CBSE board exam, so every student required special attention in mathematics.

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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**

**Latest Sample paper Class 10 maths for Term 1 2021 CBSE board**

**NCERT Solutions for Class 10 Maths Exercise 8.3 of Chapter 8 Introduction to Trigonometry**

Q1.Evaluate

Ans.

(i) sin 18°/cos 72°

Since sin θ = cos (90°-θ)

⇒sin18°=cos(90°-18°)

Substituting the value of sin 18°

cos(90°-18°)/cos 72°

cos 72°/cos 72° = 1

(ii) tan 26°/cot 64°

Since tan θ = cot (90°-θ)

⇒tan26°=cot(90°-26°)

Substituting the value of tan26°

cot(90°-26°)/cot 72°

cot 72°/cot72° = 1

(iii) cos 48° – sin 42°

Since cos θ = sin (90°-θ)

⇒cos 48°=sin(90°-48°)

Substituting the value of sin 18°

sin(90°-48°)° – sin 42°

sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

Since cosec θ = sec (90°-θ)

⇒cosec 31°=sec(90°-31°)

Substituting the value of cosec 31°

sec(90°-31°) – sec 59°

sec 59° – sec 59° = 0

**Q2.Show that:**

**(i) tan 48° tan 23° tan 42° tan 67° = 1**

**(ii) cos 38° cos 52° – sin 38° sin 52° = 0**

Ans.

(i) tan 48° tan 23° tan 42° tan 67° = 1

Since tan θ = cot (90°-θ)

⇒tan 48°=cot(90°-48°) and ⇒tan 43°=cot(90°-43°)

Substituting the value of tan 48° and tan 43°

cot(90°-48°) cot(90°-43°) tan 42° tan 67°

cot42° cot67° tan 42° tan 67°

Substituting cot42° =1/tan42° and cot67° =1/tan67°

(1/tan42°) (1/tan 67° )tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Since cos θ = sin (90°-θ)

⇒cos 38°=sin(90°-38°) and ⇒sin38°=cos(90°-38°)

Substituting the value of cos38° and sin38°

sin(90°-38°). cos 52°- cos(90°-38°).sin 52°

sin 52°.cos 52° – cos 52°sin 52° = 0

**Q3.If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A**

Ans. The given equation is

tan 2A = cot (A – 18°), where 2A is an acute angle

cot (90°-2A) = cot (A – 18°)

90°-2A = A – 18°

A + 2A =90°+ 18°=108°

3A = 108°

A = 36°

**Q4.If tan A = cot B, prove that A + B = 90°.**

Ans. The given equation is

tan A = cot B

cot (90°- A) = cot B

90°- A = B

A + B =90°,Hence proved

**Q5.If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A**

Ans.The given equation is

sec 4A = cosec (A – 20°),where 4A is an acute angle

cosec (90°- 4A) = cosec (A – 20°)

90°- 4A = A – 20°

A+4A = 90° +20°=110°

5A = 110°

A = 22°

**Q6.If A, B and C are interior angles of a triangle ABC, then show that:**

Ans.According to angle sum property of triangle

A + B + C= 180°

B + C = 180°- A

Dividing by 2

**Q7.Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.**

Ans. The given expression is

sin 61° + cos 75

cos(90° – 61°) + sin(90°-75°)

cos 29° +sin 15°

**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**

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**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 |