RBSE Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Additional Questions 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 Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Trigonometric Ratios of Acute Angles |

Exercise |
Additional Questions |

Number of Questions Solved |
31 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Additional Questions

**Multiple Choice Questions**

Question 1.

The value of sec^{2}50° – tan^{2}50° is (RBSESolutions.com) equal to:

(A) -1

(B) 0

(C) 1

(D) 2

Question 2.

Question 3.

Question 4.

Question 5.

If 3 cot φ = 2 then the (RBSESolutions.com) value of cosec^{2}φ is equal to:

(A)

(B)

(C)

(D)

**Answers**

1. C

2. A

3. C

4. D

5. B

**Very Short Answer Type Questions**

Question 1.

Write the value (RBSESolutions.com) of cosec^{2}50° – cot^{2}50°

Solution.

cosec^{2}θ – cot^{2}θ = 1

cosec^{2}50° – cot^{2}50° = 1

Question 2.

Prove that

Solution.

Question 3.

Prove that

sin θ cosec θ + cos θ sec θ = 2.

Solution.

Question 4.

Prove that

Solution.

Question 5.

Write the (RBSESolutions.com) value of (1 – cos θ)(1 + cos θ)(1 + cot^{2}θ).

Solution.

Question 6.

If cos^{2}θ – sin^{2}θ = , find the value of cos^{4}θ – sin^{4}θ

Solution.

Expression

cos^{4}θ – sin^{4}θ

Question 7.

Prove that

tan^{2}A + tan^{4}A = sec^{4}A – sin^{2}A.

Solution.

L.H.S. = tan^{2}A + tan^{4}A

= tan^{2}A (1 + tan^{2}A)

= (sec^{2}A – 1) sec^{2}A

= sec^{4}A – sec^{2}A

= R.H.S.

Question 8.

Find the (RBSESolutions.com) value of

Solution.

Question 9.

Find the value of (cos^{2}θ – 1)(1 + cot^{2}θ).

Solution.

Question 10.

Evaluate (1 – sin^{2}θ)(1 + tan^{2}θ).

Solution.

**Short Answer Type Questions**

Question 1.

Solution.

Question 2.

In any triangle ABC, if ∠B = 90° and side AB = 4 cm and AC = 5 cm, then (RBSESolutions.com) find side BC.

Solution.

Question 3.

Solution.

Question 4.

Prove that sec A (1 – sin A)(sec A + tan A) = 1

Solution.

Question 5.

Prove that

Solution.

Question 6.

Prove that

Solution.

Question 7.

Prove that

tan^{2}φ + cot^{2}φ + 2 = sec^{2}φ cosec^{2}φ

Solution.

Question 8.

If sin x + sin^{2}x = 1, show that cos^{2}x + cos^{4}x = 1.

Solution.

sin x + sin^{2}x = 1

⇒ sin x = 1 – sin^{2}x

⇒ sin x = cos^{2}x

Squaring both (RBSESolutions.com) the sides, we get

sin^{2}x = cos^{4}x

⇒ 1 – cos^{2}x = cos^{4}x

⇒ 1 = cos^{2}x + cos^{4}x

Hence proved.

Question 9.

Prove that cos^{2}θ (1 + tan^{2}θ) + sin^{2}θ (1 + cot^{2}θ) = 2

Solution.

Question 10.

If sin θ + cos θ = a and sin θ – cos θ = b then prove that a^{2} + b^{2} = 2.

Solution.

sin θ + cos θ = a …(i)

and sin θ – cos θ = b …(ii)

Squaring eqns. (i) and (ii) and (RBSESolutions.com) then adding

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

⇒ sin^{2}θ + cos^{2}θ + 2 sin θ cos θ + sin^{2}θ + cos^{2}θ – 2 sin θ cos θ = a^{2} + b^{2}

⇒ 1 + 1 = a^{2} + b^{2}

⇒ a^{2} + b^{2} = 2

Hence proved.

**Long Answer Type Questions**

Question 1.

Prove that (sec A – cos A)(cosec A – sin A) =

Solution.

Question 2.

Prove that sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ.

Solution.

Question 3.

Solution.

Question 4.

Solution.

Question 5.

Prove that sec^{6}θ = tan^{6}θ + 3 tan^{2}θ sec^{2}θ + 1.

Solution.

Question 6.

Prove that

Solution.

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

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