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

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 sec250° – tan250° 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 cosec2φ is equal to:
(A) $$\frac { \surd 3 }{ 3 }$$
(B) $$\frac { 13 }{ 9 }$$
(C) $$\frac { 9 }{ 13 }$$
(D) $$\frac { 3 }{ \surd 3 }$$

1. C
2. A
3. C
4. D
5. B

Question 1.
Write the value (RBSESolutions.com) of cosec250° – cot250°
Solution.
cosec2θ – cot2θ = 1
cosec250° – cot250° = 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 + cot2θ).
Solution.

Question 6.
If cos2θ – sin2θ = $$\frac { 1 }{ 2 }$$ , find the value of cos4θ – sin4θ
Solution.
Expression
cos4θ – sin4θ

Question 7.
Prove that
tan2A + tan4A = sec4A – sin2A.
Solution.
L.H.S. = tan2A + tan4A
= tan2A (1 + tan2A)
= (sec2A – 1) sec2A
= sec4A – sec2A
= R.H.S.

Question 8.
Find the (RBSESolutions.com) value of

Solution.

Question 9.
Find the value of (cos2θ – 1)(1 + cot2θ).
Solution.

Question 10.
Evaluate (1 – sin2θ)(1 + tan2θ).
Solution.

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
tan2φ + cot2φ + 2 = sec2φ cosec2φ
Solution.

Question 8.
If sin x + sin2x = 1, show that cos2x + cos4x = 1.
Solution.
sin x + sin2x = 1
⇒ sin x = 1 – sin2x
⇒ sin x = cos2x
Squaring both (RBSESolutions.com) the sides, we get
sin2x = cos4x
⇒ 1 – cos2x = cos4x
⇒ 1 = cos2x + cos4x
Hence proved.

Question 9.
Prove that cos2θ (1 + tan2θ) + sin2θ (1 + cot2θ) = 2
Solution.

Question 10.
If sin θ + cos θ = a and sin θ – cos θ = b then prove that a2 + b2 = 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 = a2 + b2
⇒ sin2θ + cos2θ + 2 sin θ cos θ + sin2θ + cos2θ – 2 sin θ cos θ = a2 + b2
⇒ 1 + 1 = a2 + b2
⇒ a2 + b2 = 2
Hence proved.

Question 1.
Prove that (sec A – cos A)(cosec A – sin A) = $$\frac { 1 }{ tanA+cotA }$$
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 sec6θ = tan6θ + 3 tan2θ sec2θ + 1.
Solution.

Question 6.
Prove that

Solution.

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