sin2θ = 2sinθ - 1, (0° < θ < 90°)
find the value of \(\frac{1+cosecθ}{1-cosθ}\)
Answer & explanation
Correct answer: option 2
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