RBSE Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Trigonometric Ratios of Acute Angles |

Exercise |
Exercise 14.3 |

Number of Questions Solved |
14 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.3

Question 1.

Prove (RBSESolutions.com) that cos θ tan θ = sin θ.

Solution.

L.H.S. = cos θ tan θ

= cos θ x

= sin θ

= R.H.S.

Hence proved.

Question 2.

Prove that (1 – sin^{2}θ) tan^{2}θ = sin^{2}θ.

Solution.

Question 3.

Prove (RBSESolutions.com) that

Solution.

Question 4.

Prove that

(sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = 2.

Solution.

Question 5.

Prove that

cosec^{6}θ – cot^{6}θ = 1 + 3 cosec^{2}θ cot^{2}θ.

Solution.

Question 6.

Prove that (RBSESolutions.com)

sin^{2}θ cos θ + tan θ sin θ + cos^{3}θ = sec θ.

Solution.

Question 7.

Prove that

Solution.

Question 8.

Prove (RBSESolutions.com) that

Solution.

Question 9.

Prove that

Solution.

Question 10.

Prove that

Solution.

Question 11.

Prove that

Solution.

Question 12.

Prove that

Solution.

Question 13.

Prove that

Solution.

Question 14.

Prove that (1 + cot θ – cosec θ)(1 + tan θ + sec θ) = 2

Solution.

We hope the given RBSE Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.3 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.3, drop a comment below and we will get back to you at the earliest.

## Leave a Reply