If tan2θ – 3sec θ + 3 = 0, 0°<θ<90°, then the value of sin θ + cot θ is:
Show answer & explanation
Correct answer: D
Explanation
Correct Answer is option 4 i.e. 5√3/6
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Concept |
Understanding |
Application |
Fair calculation |
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Tanθ = Sinθ/Cosθ Secθ = 1/Cosθ Sin2 θ + Cos2θ = 1 |
Given tan2θ – 3 sec θ + 3 = 0 0 < θ < 90o
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tan2θ – 3 sec θ + 3 = 0 ⇒ sin2θ/Cos2θ – 3/Cosθ + 3 = 0 ⇒ sin2θ – 3cos θ + 3cos2θ=0 ⇒ 1- 3cosθ + 2cos2θ=0 ⇒ cosθ = 1 or ½ we take cos θ= ½ because 0 < θ < 90 Hence, θ = 60o Now, sin θ + cot θ ⇒ sin60 + cot60 ⇒ √3/2 + 1/√3 ⇒ 5√3/6 |
√3/2 + 1/√3 = (3 + 2)/2√3 = 5/2√3 = 5√3/6 |