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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3
NCERT TEXTBOOK EXERCISES
Question 1.

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:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)
= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° .$\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}$
= 1 = RHS
(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin 38° sin (90° – 38°)
= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS
Question 3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = 108°/3
∴ ∠ A = 36°
Question 4. If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B ⇒ tan A = tan (90° – B) [ ∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B ⇒ A + B = 90° Proved
Question 5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°) [cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20° ⇒ 5A = 110°
A = 110°/5
A = 22°
∴ ∠ A = 22°
Question 6. If A, Band Care interior angles of a triangle ABC, then show that: sin $\frac { B+C }{ 2 }$ = cos A/2
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°
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 PDF
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