If 3sin θ = 2cos2θ, 0° < θ < 90°, then the value of (tan2θ + sec2θ – cosec2θ) is?
Show answer & explanation
Correct answer: B
Explanation
Correct Answer is option 2 i.e. -7/3
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Concept |
Understanding |
Application |
Fair calculation |
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Value putting method. |
Given 3sinθ = 2Cos2θ We need to evaluate what value of θ will satisfy this where 0 <θ< 90 |
Putting θ = 30 3 × 1/2 = 2 × (√3/2)2 ⇒ 3/2 = 3/2 Satisfies the equation. Hence, θ = 30 Now, (tan2θ + sec2θ - cosec2θ) ⇒(tan230 + sec230 - cosec230) ⇒[(1/√3)2 + (2/√3)2–(2)2] ⇒ -7/3 |
[(1/√3)2 + (2/√3)2 – (2)2] ⇒ [1/3 + 4/3 – 4] ⇒ 5/3 – 4 ⇒ -7/3
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