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
Ex 8.3 Class 10 Maths Question 1.
Evaluate:
Solution:
Ex 8.3 Class 10 Maths 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:
Ex 8.3 Class 10 Maths 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 = \(\frac { 108 }{ 3 } \) = 36°.
Ex 8.3 Class 10 Maths 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.
Ex 8.3 Class 10 Maths 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 ⇒ \(\frac { 110 }{ 5 } \) = A
⇒ A = 22°
Ex 8.3 Class 10 Maths Question 6.
If A, B and C are interior angles of a triangle ABC, then show that: sin (\(\frac { B+C }{ 2 }\)) = cos \(\frac { A }{ 2 }\)
Solution:
Ex 8.3 Class 10 Maths 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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