sin2θ = 2sinθ - 1, (0° < θ < 90°) find the value of \(\frac{1+cosecθ}{1-cosθ}\) |
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In case of sin and cos try to put θ = 90° or 0° Put θ = 90° ⇒ sin2θ = 2sinθ - 1 1 = 2 - 1 1 = 1 satisfied Put θ = 90°, ⇒\(\frac{1+cosecθ}{1-cosθ}\) = \(\frac{1+1}{1-0}\)=2 |