NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3.

• Introduction to Trigonometry Class 10 Ex 8.1
• Introduction to Trigonometry Class 10 Ex 8.2
• Introduction to Trigonometry Class 10 Ex 8.3
• Introduction to Trigonometry Class 10 Ex 8.4

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 8
Chapter Name	Introduction to Trigonometry
Exercise	Ex 8.3
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:

Solution:

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
We have    tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [ ∵ cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18°
⇒ 2A + A = 90° + 18°
⇒ 3A = 108°
⇒ A = = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
We have:    tan A = cot B
⇒ cot  (90° – A) = cot B
⇒ 90° – A = B    ⇒ A + B = 90°
Hence,  Proved.

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
We have     sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)   [∵  cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20°   ⇒ 90° + 20° = 5A
⇒ 110° = 5A    ⇒    = A
⇒ A = 22°

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that: sin () = cos
Solution:

Question 7.
Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°) = cos 23° + sin 15°.

 

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