Question
If 3 sec² + tanθ – 7 = 0,0° <
θ < 90°, then what is the value of (cosec2θ + cosθ)/(sin2θ + cotθ) ?Solution
3 sec2 θ + tan θ = 7 ⇒ Put θ = 45 ⇒ 3 sec2 45 + tan 45 = 7 ⇒ 3 × 2 + 1 = 7 ⇒ 7 = 7 (satisfied) Now, ⇒ (cosec 2θ + cos θ)/(sin2θ + cot θ) ⇒ (cosec 90 + cos 45)/(sin 90 + cot 45) ⇒ (1 + 1/√2)/(1 + 1) ⇒ (√2 + 1)/2√2 ⇒ (2 + √2)/4
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