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

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.

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 sin²30° + cos²30° is:
(A) 0
(B) 2
(C) 3
(D) 1

Question 8.
The value of cosec²55° – cot²55° 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 cos⁴θ – sin⁴θ = 1 – 2 sin²θ.
Solution.
L.H.S. = cos⁴θ – sin⁴θ
= (cos²θ)² – (sin²θ)²
= (cos²θ – sin²θ)(cos²θ + sin²θ)
= (cos²θ – sin²θ) (1)
= (1 – sin²θ – sin²θ)
= 1 – 2 sin²θ
= R.H.S.
Hence proved.

Question 31.
Prove that sec²θ – cosec²θ = tan²θ – cot²θ.
Solution.
L.H.S. = sec²θ – cosec²θ
= 1 + tan²θ – (1 + cot²θ)
= 1 + tan²θ – 1 – cot²θ
= tan²θ – cot²θ
= R.H.S.
Hence proved.

Question 32.

Solution.

Question 33.
Prove that
(sin A + cosec A)² + (cos A + sec A)² = tan²A + cot²A + 7.
Solution.
L.H.S.
= (sin A + cosec A)² + (cos A + sec A)²
= sin²A + cosec²A + 2 sin A cosec A + cos²A + sec²A + 2 cos A sec A

Question 34.

Solution.

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