NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3.

- Introduction to Trigonometry Class 10 Ex 8.1
- Introduction to Trigonometry Class 10 Ex 8.2
- Introduction to Trigonometry Class 10 Ex 8.3
- Introduction to Trigonometry Class 10 Ex 8.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Introduction to Trigonometry |

Exercise |
Ex 8.3 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

**Ex 8.3 Class 10 Maths Question 1.**

**Evaluate:**

**Solution:**

**Ex 8.3 Class 10 Maths Question 2.**

**Show that:**

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

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

**Solution:**

**Ex 8.3 Class 10 Maths Question 3.**

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

**Solution:**

We have tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°) [ ∵ cot (90° – θ) = tan θ]

⇒ 90° – 2A = A – 18°

⇒ 2A + A = 90° + 18°

⇒ 3A = 108°

⇒ A = \(\frac { 108 }{ 3 } \) = 36°.

**Ex 8.3 Class 10 Maths Question 4.**

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

**Solution:**

We have: tan A = cot B

⇒ cot (90° – A) = cot B

⇒ 90° – A = B ⇒ A + B = 90°

Hence, Proved.

**Ex 8.3 Class 10 Maths Question 5.**

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

**Solution:**

We have sec 4A = cosec (A – 20°)

⇒ cosec (90° – 4A) = cosec (A – 20°) [∵ cosec (90° – θ) = sec θ]

⇒ 90° – 4A = A – 20° ⇒ 90° + 20° = 5A

⇒ 110° = 5A ⇒ \(\frac { 110 }{ 5 } \) = A

⇒ A = 22°

**Ex 8.3 Class 10 Maths Question 6.**

**If A, B and C are interior angles of a triangle ABC, then show that: sin (\(\frac { B+C }{ 2 }\)) = cos \(\frac { A }{ 2 }\)**

**Solution:**

**Ex 8.3 Class 10 Maths Question 7.**

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

**Solution:**

sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°) = cos 23° + sin 15°.

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