Ncert Solutions Class 10 Maths Trigonometry Ex 8.3
NCERT solutions for class-10 maths Chapter-8 Introduction of Trigonometry is prepared by academic team of entrancei. we have prepared solutions for all exercises of this chapter. Given below is step by step solutions of all questions given in NCERT textbook for chapter-8. You have learned the formula of the given chapter. Entrancei prepared a detail notes and additional questions for class-10 maths with short notes of all maths formula of class-10 maths. These NCERT solutions will be updated according to the latest 2020-21 syllabus. The concept and formulas for each question have also been updated.
1. Evaluate :
(i) sin 18°/cos 72° (ii) tan 26°/cot 64° (iii) cos 48° – sin 42° (iv) cosec 31° – sec 59°
Answer:
(i) sin 18°/cos 72°
= sin (90° - 18°) /cos 72°
= cos 72° /cos 72° = 1
(ii) tan 26°/cot 64°
= tan (90° - 36°)/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
Answer:
(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°) = 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.
Answer:
tan 2A = cot (A- 18°)
⇒ cot (90° - 2A) = cot (A -18°)
Equating angles,
⇒ 90° - 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°
4. If tan A = cot B, prove that A + B = 90°.
Answer:
tan A = cot B
⇒ tan A = tan (90° - B)
⇒ A = 90° - B
⇒ A + B = 90°
5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Answer:
sec 4A = cosec (A - 20°)
⇒ cosec (90° - 4A) = cosec (A - 20°)
Equating angles,
90° - 4A= A- 20°
⇒ 110° = 5A
⇒ A = 22°
6. If A, B and C are interior angles of a triangle ABC, then show that
sin (B+C/2) = cos A/2
Answer:
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
sin 67° + cos 75°
= sin (90° - 23°) + cos (90° - 15°)
= cos 23° + sin 15°
Ncert Solutions Class 10 Maths Trigonometry Ex 8.3
Source: https://www.entrancei.com/ncert-solutions-class-10-maths-chapter-8-exercise-8.3