RBSE Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Miscellaneous Exercise 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 Miscellaneous Exercise.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Trigonometric Ratios of Acute Angles |

Exercise |
Miscellaneous Exercise |

Number of Questions Solved |
34 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Miscellaneous Exercise

**Multiple Choice Questions**

Question 1.

If tan θ = √3, then the value (RBSESolutions.com) of sinθ is:

(A) \(\frac { 1 }{ \surd 3 }\)

(B) \(\frac { \surd 3 }{ 2 }\)

(C) \(\frac { 2 }{ \surd 3 }\)

(D) 1

Question 2.

If sin θ = \(\frac { 5 }{ 13 }\), then the value of tan θ is:

(A) \(\frac { 5 }{ 12 }\)

(B) \(\frac { 12 }{ 13 }\)

(C) \(\frac { 13 }{ 12 }\)

(D) \(\frac { 12 }{ 5 }\)

Question 3.

If √3 cos A = sin A, then the (RBSESolutions.com) value of cot A is:

(A) √3

(B) 1

(C) \(\frac { 1 }{ \surd 3 }\)

(D) 2

Question 4.

In figure, the value of cot A is:

(A) \(\frac { 12 }{ 13 }\)

(B) \(\frac { 5 }{ 12 }\)

(C) \(\frac { 5 }{ 13 }\)

(D) \(\frac { 13 }{ 5 }\)

Question 5.

In figure, the value (RBSESolutions.com) of tan θ is:

(A) 2

(B) \(\frac { 1 }{ \surd 5 }\)

(C) \(\frac { 2 }{ \surd 5 }\)

(D) \(\frac { 1 }{ 2 }\)

Question 6.

In figure, the value of cosec α is:

(A) \(\frac { y }{ x }\)

(B) \(\frac { y }{ z }\)

(C) \(\frac { x }{ z }\)

(D) \(\frac { x }{ y }\)

Question 7.

The value of sin^{2}30° + cos^{2}30° is:

(A) 0

(B) 2

(C) 3

(D) 1

Question 8.

The value of cosec^{2}55° – cot^{2}55° is:

(A) 1

(B) 2

(C) 3

(D) 0

Question 9.

If cot φ = \(\frac { 20 }{ 21 }\) then the (RBSESolutions.com) value of cosec φ is:

(A) \(\frac { 21 }{ 20 }\)

(B) \(\frac { 20 }{ 29 }\)

(C) \(\frac { 29 }{ 21 }\)

(D) \(\frac { 21 }{ 29 }\)

Question 10.

If in ∆ABC, ∠B = 90°, c = 12 cm and a = 9 cm then the value of cos C is:

(A) \(\frac { 3 }{ 5 }\)

(B) \(\frac { 3 }{ 4 }\)

(C) \(\frac { 5 }{ 3 }\)

(D) \(\frac { 4 }{ 5 }\)

Question 11.

The value of: (sec 40° + tan 40°)(sec 40° – tan 40°) is equal to:

(A) – 1

(B) 1

(C) cos 40°

(D) sin 40°

Question 12.

Question 13.

Question 14.

Answers

1. B

2. A

3. C

4. B

5. D

6. B

7. D

8. A

9. C

10. A

11. B

12. A

13. C

14. D

Question 15.

If cosec θ = \(\frac { 41 }{ 40 }\), then find the (RBSESolutions.com) value of tan θ and cos θ.

Solution.

Question 16.

In an ∆ABC, if ∠B = 90° and AB = 12 cm and BC = 5 cm, then find (RBSESolutions.com) the value of sin A, tan A, sin C and cot C.

Solution.

∆ABC is a right angled triangle right angled at B.

Question 17.

Solution.

Question 18.

Solution.

Question 19.

If cot A = √3, then (RBSESolutions.com) prove that sin A cos B + cos A sin B = 1.

Solution.

Question 20.

Solution.

Question 21.

Solution.

Question 22.

Solution.

Question 23.

Solution.

Question 24.

If sin A = \(\frac { 1 }{ 3 }\), then (RBSESolutions.com) evaluate cos A . cosec A + tan A . sec A.

Solution.

Question 25.

Solution.

Question 26.

Solution.

Question 27.

Solution.

Question 28.

Solution.

Question 29.

Solution.

Question 30.

Prove that cos^{4}θ – sin^{4}θ = 1 – 2 sin^{2}θ.

Solution.

L.H.S. = cos^{4}θ – sin^{4}θ

= (cos^{2}θ)^{2} – (sin^{2}θ)^{2}

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

= (cos^{2}θ – sin^{2}θ) (1)

= (1 – sin^{2}θ – sin^{2}θ)

= 1 – 2 sin^{2}θ

= R.H.S.

Hence proved.

Question 31.

Prove that sec^{2}θ – cosec^{2}θ = tan^{2}θ – cot^{2}θ.

Solution.

L.H.S. = sec^{2}θ – cosec^{2}θ

= 1 + tan^{2}θ – (1 + cot^{2}θ)

= 1 + tan^{2}θ – 1 – cot^{2}θ

= tan^{2}θ – cot^{2}θ

= R.H.S.

Hence proved.

Question 32.

Solution.

Question 33.

Prove that

(sin A + cosec A)^{2} + (cos A + sec A)^{2} = tan^{2}A + cot^{2}A + 7.

Solution.

L.H.S.

= (sin A + cosec A)^{2} + (cos A + sec A)^{2}

= sin^{2}A + cosec^{2}A + 2 sin A cosec A + cos^{2}A + sec^{2}A + 2 cos A sec A

Question 34.

Solution.

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