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

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  $$\frac { sin\theta }{ cos\theta }$$
= sin θ
= R.H.S.
Hence proved.

Question 2.
Prove that (1 – sin2θ) tan2θ = sin2θ.
Solution.

Question 3.
Prove (RBSESolutions.com) that $$\frac { { cos }^{ 2 }\theta }{ sin\theta } +sin\theta =cosec\theta$$
Solution.

Question 4.
Prove that
(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2.
Solution.

Question 5.
Prove that
cosec6θ – cot6θ = 1 + 3 cosec2θ cot2θ.
Solution.

Question 6.
Prove that (RBSESolutions.com)
sin2θ cos θ + tan θ sin θ + cos3θ = sec θ.
Solution.

Question 7.
Prove that $$\frac { cos\theta }{ 1-tan\theta } +\frac { sin\theta }{ 1-cot\theta } =sin\theta +cos\theta$$
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.

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