If 12cot2θ - 31cosecθ + 32 = 0, 0° < θ < 90°, then the values of tanθ will be:
Show answer & explanation
Correct answer: B
Explanation
Correct Answer is option 2 i.e. 4/3, 3√7/7
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Concept |
Understanding |
Application |
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cot2θ = cosec2θ – 1 cos2θ = 1 – sin2θ |
Given 12cot2θ – 31cosecθ + 32 = 0 0° < θ < 90° |
12cot2θ – 31cosecθ + 32 = 0 ⇒ 12(cosec2θ – 1) – 31cosecθ + 32 = 0 ⇒ 12cosec2θ – 31cosecθ + 20 = 0 ⇒ 12cosec2θ – 16cosecθ – 15cosecθ + 20 = 0 ⇒ 4cosecθ(3cosec θ – 4) – 5(3cosec θ – 4) = 0 ⇒ (4cosecθ – 5) (3cosec θ – 4) = 0 ⇒ cosec θ = 5/4 or 4/3 ⇒ sin θ = 4/5 or 3/4 Now, cos2θ = 1 – (4/5)2 = 9/25 or 1 – (3/4)2 = 7/16 ⇒ cos θ = 3/5 or √7/4 ⇒ tan θ = sin θ/cos θ ⇒ tan θ = (4/5)/(3/5) or (3/4)/(√7/4) ⇒ tan θ = 4/3 or 3√7/7 |