J's answer to help's Secondary 4 A Maths Singapore question.
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(2 - cos θ)(tan² θ - 1) = 0, 0 ≤ θ ≤ π
(2 - cos θ)(tan θ + 1)(tan θ - 1) = 0
(Recall the factorisation property a² - b² = (a + b)(a - b)
2 - cos θ = 0 or tan θ + 1 = 0 or tan θ - 1 = 0
cos θ = 2 (rejected as -1 cos θ ≤ 1) or tan θ = -1 ( negative, look at 2nd and 4th quadrant) or tan θ = 1 (positive, look at 1st and 3rd quadrant)
θ = tan-¹ (-1) or tan θ = tan-¹ (1)
Basic angle = -π/4 or basic angle = π/4
Since 0 ≤ θ ≤ π,
θ = -π/4 + π or θ = π/4
θ = 3π/4 or θ = π/4
(2 - cos θ)(tan θ + 1)(tan θ - 1) = 0
(Recall the factorisation property a² - b² = (a + b)(a - b)
2 - cos θ = 0 or tan θ + 1 = 0 or tan θ - 1 = 0
cos θ = 2 (rejected as -1 cos θ ≤ 1) or tan θ = -1 ( negative, look at 2nd and 4th quadrant) or tan θ = 1 (positive, look at 1st and 3rd quadrant)
θ = tan-¹ (-1) or tan θ = tan-¹ (1)
Basic angle = -π/4 or basic angle = π/4
Since 0 ≤ θ ≤ π,
θ = -π/4 + π or θ = π/4
θ = 3π/4 or θ = π/4
Date Posted:
3 years ago
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