NCERT Solutions Class 10 Maths
Chapter – 8 (Introduction to Trigonometry)
The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 8 Introduction to Trigonometry Exercise 8.3 has been provided here to help the students in solving the questions from this exercise.
Chapter : 8 Introduction to Trigonometry
- NCERT Class 10 Maths Solution Ex – 8.1
- NCERT Class 10 Maths Solution Ex – 8.2
- NCERT Class 10 Maths Solution Ex – 8.4
Exercise – 8.3
1. Evaluate :
(i) sin 18°/cos 72°
(ii) tan 26°/cot 64°
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Solution –
(i) sin 18°/cos 72°
= sin (90° – 18°) /cos 72°
= cos 72° /cos 72°
= 1
(ii) tan 26°/cot 64°
= tan (90° – 26°)/cot 64°
= cot 64°/cot 64°
= 1
(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42°
= 0
(iv) cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59°
= sec 59° – sec 59°
= 0
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) tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
(Since, tan A° cot A° = 1)
= 1 × 1
= 1
(ii) cos 38° cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52°
= 0
3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution –
tan 2A = cot (A – 18°)
We know that tan 2A = cot (90° – 2A)
⇒ cot (90° – 2A) = cot (A -18°)
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
⇒ A = 108°/3
⇒ A = 36°
Therefore, the value of A = 36°
4. If tan A = cot B, prove that A + B = 90°.
Solution –
tan A = cot B
We know that cot B = tan (90° – B)
To prove A + B = 90°,
tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°
Hence Proved.
5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution –
sec 4A = cosec (A – 20°)
We know that sec 4A = cosec (90° – 4A)
⇒ cosec (90° – 4A) = cosec (A – 20°)
⇒ 90° – 4A= A – 20°
⇒ 110° = 5A
⇒ A = 110°/ 5
⇒ A = 22°
Therefore, the value of A = 22°
6. If A, B and C are interior angles of a triangle ABC, then show that
Solution – In a triangle, sum of all the interior angles
A + B + C = 180°
⇒ B + C = 180° – A
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = (90° – A/2)
⇒ sin (B + C)/2 = sin (90° – A/2)
⇒ sin (B + C)/2 = cos A/2
Hence proved.
7. Express sin 67° + 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°
