Exercise 19C | Q.1, 2 & 3 | Chapter-19 | Trigonometric Ratios | Class 9 | ICSE

Описание: 📘 Exercise 19C | Q.1, 2 & 3 | Trigonometric Ratios | Class 9 | ICSE Mathematics

In this video, we solve Exercise 19C (Questions 1, 2 & 3) from Chapter 19 – Trigonometric Ratios step by step.
Each solution is explained clearly using ICSE Class 9 Mathematics textbook concepts, helping you build strong fundamentals in Trigonometry.

🎯 What You’ll Learn in This Video:
✅ Trigonometric identities and values of standard angles
✅ Relationship between sine, cosine, and tangent ratios
✅ Step-by-step solutions of ICSE Board-style questions
✅ Concept clarity with logic behind each formula


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📖 Questions Covered in this Video:

Q.1
(i) Prove that sin 60° = 2 sin 30° cos 30°.
(ii) Without using trigonometric tables, find the value of 3 sin² 45° + 2 cos² 60°.

Q.2
If 0° ≤ x ≤ 90°, state the numerical value of x for which sin x° = cos x°.

Q.3 Find the value of:
(i) sin² 60° + cos² 45°
(ii) 3 cos² 30° + tan² 60°
(iii) 4 sin² 60° + 3 tan² 30° – 8 sin 45° cos 45°
(iv) 2 sin² 30° – 3 cos² 45° + tan² 60°
(v) (sin 60° / cos² 45°) – 3 tan 30° + 5 cos 90°
(vi) cos 90° + cos² 45° sin 30° tan 45°
(vii) cos² 45° + sin² 60° + sin² 30°
(viii) sin² 30° + cos² 60°
(ix) (sin² 45° + cos² 45°) / tan² 60°
(x) (5 sin² 30° + cos² 45° – 4 tan² 30°) / (2 sin 30° cos 30° + tan 45°)
(xi) 2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
(xii) (4/3) tan² 30° + sin² 60° – 3 cos² 60° + (3/4) tan² 60° – 2 tan² 45°
(xiii) (cos 0° + sin 45° + sin 30°) × (sin 90° – cos 45° + cos 60°)


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📕 Book Credit: ICSE Mathematics – Class IX
© Council for the Indian School Certificate Examinations (CISCE)

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