Find the acute angle θ such that 5tan2 θ + 3 = 9sec θ.

1.	Find the acute angle θ such that 5tan2 θ + 3 = 9sec θ.
Answer» 5tan θ + 3 = 9sec θ ∴ 5(sec² θ – 1) + 3 = 9sec θ ∴ 5sec² θ – 5 + 3 = 9sec θ ∴ 5sec² θ – 9sec θ – 2 = 0 ∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0 ∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0 ∴ (sec θ – 2) (5sec θ + 1) = 0 ∴ sec θ – 2 = 0 or 5sec θ + 1 = 0 ∴ sec θ = 2 or sec θ = -1/5 Since sec θ ≥ 1 or sec θ ≤ -1, sec θ = 2 ∴ θ = 60° … [ ∵ sec 60° = 2]

Answer»

5tan θ + 3 = 9sec θ

∴ 5(sec² θ – 1) + 3 = 9sec θ

∴ 5sec² θ – 5 + 3 = 9sec θ

∴ 5sec² θ – 9sec θ – 2 = 0

∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0

∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0

∴ (sec θ – 2) (5sec θ + 1) = 0

∴ sec θ – 2 = 0 or 5sec θ + 1 = 0

∴ sec θ = 2 or sec θ = -1/5

Since sec θ ≥ 1 or sec θ ≤ -1,

sec θ = 2

∴ θ = 60° … [ ∵ sec 60° = 2]

Discussion