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Question
Find the acute angle θ such that 5tan2θ + 3 = 9secθ.
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Solution
∴ 5 tan2θ+ 3 = 9 secθ
∴ 5 (sec2θ – 1) + 3 = 9 secθ
∴ 5 sec2θ – 5 + 3 = 9 secθ
∴ 5 sec2θ – 9secθ − 2 = 0
∴ 5 sec2θ – 10secθ + secθ – 2 = 0
∴ 5 secθ (secθ – 2) + 1 (secθ – 2) = 0
∴ (secθ – 2)(5secθ + 1) = 0
∴ secθ – 2 = 0 or 5secθ + 1 = 0
∴ secθ = 2 or secθ = `-1/5`
But secθ ≥ 1 or secθ ≤ – 1 for all θ ∈ R, where cosθ ≠ 0
∴ secθ ≠ `-1/5`
∴ secθ = 2 = sec60° ...[∵ θ is acute angle]
∴ θ = 60°
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