NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 8 **Introduction to Trigonometry Class 10 NCERT Solutions 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 8.4

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

NCERT Solutions for Class 10 Maths

**Page No : 189**

**Question 1.** Evaluate :

(i) \(\frac { { sin18 }^{ 0 } }{ { cos72 }^{ 0 } } \)

(ii) \(\frac { { tan26 }^{ 0 } }{ { cot64 }^{ 0 } } \)

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

**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: **

**Question 3.** If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

**Solution: **Given that

tan 2A = cot (A – 18

^{0})

cot (90

^{0}– 2A) = cot (A -18

^{0})

90

^{0}– 2A = A – 18

^{0}

108

^{0}= 3A

A = 36

^{0}

**Question 4.** If tan A = cot B, prove that A + B = 90°.

**Solution: **Given that

tan A = cot B

tan A = tan (90

^{0}– B)

A = 90

^{0}– B

A + B = 90

^{0}

**Question 5.** If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

**Solution: **Given that

Sec 4A = cosec (A – 20

^{0})

Cosec (90

^{0}– 4A) = cosec (A – 20

^{0})

90

^{0 }– 4A = A – 20

^{0}

110

^{0}= 5A

A = 22

^{0}

Page No : 190

**Question 6.** If A, B and C are interior angles of a triangle ABC, then

Show that \(sin\left( \frac { B+C }{ 2 } \right) =cos\frac { A }{ 2 } \)

**Solution: **We know that for a triangle ∆ABC

∠A + ∠B + ∠C = 180:

∠B + ∠C = 180 – ∠A

\(\frac { \angle B+\angle C }{ 2 } ={ 90 }^{ 0 }-\frac { \angle A }{ 2 } \)

\(sin\left( \frac { B+C }{ 2 } \right) =sin\left( { 90 }^{ 0 }-\frac { A }{ 2 } \right) \)

\(=cos\left( \frac { A }{ 2 } \right) \)

**Question 7.** Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

**Solution: **sin 67

^{0}+ cos 75

^{0}

= sin (90

^{0}– 23

^{0}) + cos (90

^{0}– 15

^{0})

= cos 23

^{0}+ sin 15

^{0}

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