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If sin2θ = 2 sinθ - 1 (0° ≤ θ ≤ 90°), find the value of \(\frac{1 + cosecθ + cosθ}{1 - \frac{2cosθ}{3} - cosθ}\) |
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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 |
