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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