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If secθ + tanθ = 2 + √5 , then the value of sinθ is (0° ≤ θ ≤ 90°)
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√3 2 -
2 √5 -
1 √5 -
4 5
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Correct Option: B
sec θ + tan θ = 2 + 5
∵ sec²θ – tan²θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1
| ⇒ sec θ – tan θ = | √5 + 2 |
| = | × | = | √5 + 2 | √5 - 2 | 5 - 4 |
= √5 - 2
∴ sec θ + tan θ + sec θ – tan θ = 2 + √5 + √5 - 2
⇒ 2 secθ = 2√5
⇒ secθ = √5 ....(i)
Again,
sec θ + tan θ – (sec θ – tan θ)
= 2 + √5 - √5 + 2
⇒ 2 tanθ = 4
⇒ tanθ = 2 ....(ii)
| ∴ sin θ | = | sec θ | √5 |
