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secondary 4 | A Maths
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I need help with part (iii), thanks!
Date Posted: 3 years ago
Views: 179
We need to use the property sec² θ = 1 + tan² θ
sec² 10° = 1 + tan² 10°
1 / cos² 10° = 1 + tan² 10°
1 / (cos 10°)² - 1 = tan² 10°
1 / (sin(90°-10°))² - 1 = tan² 10°
1 / (sin 80°)² - 1 = tan² 10°
1 / (sin 100°)² - 1 = tan² 10°
(From part i you would already have found that sin 100° = sin(180° - 100°) = sin 80°)
1/k² - 1 = tan² 10°
tan 10° = √(1/k² - 1) or √((k²-1)/k²)
sec² 10° = 1 + tan² 10°
1 / cos² 10° = 1 + tan² 10°
1 / (cos 10°)² - 1 = tan² 10°
1 / (sin(90°-10°))² - 1 = tan² 10°
1 / (sin 80°)² - 1 = tan² 10°
1 / (sin 100°)² - 1 = tan² 10°
(From part i you would already have found that sin 100° = sin(180° - 100°) = sin 80°)
1/k² - 1 = tan² 10°
tan 10° = √(1/k² - 1) or √((k²-1)/k²)
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