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.

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

**Question 1.**

**Solution:**

**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:**

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

= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)

= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° .\(\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)

= 1 = RHS

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

= cos 38° cos (90° – 38°) – sin 38° sin (90° – 38°)

= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS

**Question 3.**

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

**Solution:**

tan 2A = cot (A – 18°)

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

⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = \(\frac { 108° }{ 3 }\)

∴ ∠ A = 36°

**Question 4.**

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

**Solution:**

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

⇒ A = 90° – B ⇒ A + B = 90° Proved

**Question 5.**

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

**Solution:**

sec 4A = cosec (A – 20°)

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

⇒ 90° – 4A = A – 20° ⇒ 5A = 110°

A = \(\frac { 110° }{ 5 }\)

A = 22°

∴ ∠ A = 22°

**Question 6.**

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

**Solution:**

**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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