If 2cos²θ + 5sinθ = 4 where 0° < θ < 90°, what is the value of cotθ ?

Question: If 2cos²θ + 5sinθ = 4 where 0° < θ < 90°, what is the value of cotθ ?

Solution:
দেওয়া আছে,
2cos²θ + 5sinθ = 4
⇒ 2(1 - sin²θ) + 5sinθ = 4 [cos²θ = 1 - sin²θ]
⇒ 2 - 2sin²θ + 5sinθ - 4 = 0
⇒ - 2sin²θ + 5sinθ - 2 = 0
⇒ 2sin²θ - 5sinθ + 2 = 0
⇒ 2sin²θ - 4sinθ - sinθ + 2 = 0
⇒ 2sinθ(sinθ - 2) -1(sinθ - 2) = 0
⇒ (sinθ - 2)(2sinθ - 1) = 0

যেহেতু sinθ এর সর্বোচ্চ মান 1, তাই sinθ = 2 অসম্ভব।
∴ 2sinθ - 1=0
⇒ sinθ = 1/2
⇒ sinθ = sin30°
∴ θ = 30°

অতএব, cotθ = cot30° = √3