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Find the acute angle θ such that 5tan2θ + 3 = 9secθ. - Mathematics and Statistics

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प्रश्न

Find the acute angle θ such that 5tan2θ + 3 = 9secθ.

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उत्तर

∴ 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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Fundamental Identities
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पाठ 2: Trigonometry - 1 - EXERCISE 2.2 [पृष्ठ ३१]

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