If sin2θ = 2 sinθ - 1 (0° ≤ θ ≤ 90°), find the value of \(\frac{1 + cosecθ + cosθ}{1 - \frac{2cosθ}{3} - cosθ}\)
Answer & explanation
Correct answer: option 3
sin2θ = 2 sinθ - 1
sinθ = 2 - \(\frac{1}{sinθ}\)
sinθ + \(\frac{1}{sinθ}\) = 2
then θ° = 90°
Now put and find
\(\frac{1 + cosecθ + cosθ}{1 - \frac{2cosθ}{3} - cosθ}\) = \(\frac{1 + 1 + 0}{1 - 0 - 0}\) = 2