If Cotθ + tanθ = 2Secθ, 0o < θ < 90o, then the value of (tan2θ - Secθ)/(Cot2θ + Cosecθ) is:
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Correct answer: B
Explanation
The correct answer is option 2 i.e (2√3 - 1)/11
⇒ Cotθ + tanθ = 2Secθ
⇒ (cosθ / sinθ) + (sinθ / cosθ) = 2secθ
⇒ (cos2θ + sin2θ) / cosθsinθ = 2secθ
⇒ 1 / sinθcosθ = 2secθ
⇒ secθcosecθ = 2secθ
⇒ cosecθ = 2
⇒ θ = 30o (as 0o< θ < 90o)
⇒ (tan2θ - Secθ) / (Cot2θ + Cosecθ)
⇒ (tan60o - Sec30o) / (Cot60o + Cosec30o)
⇒ (√3 - 2 / √3) / (1 / √3 + 2)
⇒ (1 / √3) / (1 + 2√3 / √3)
⇒ 1 / (1 + 2√3)
By rationalizing the denominator
⇒ (2√3 - 1) / 11