If $\sin ^2 \theta=2 \sin \theta-1, 0^{\circ} \leq \theta \leq 90^{\circ}$, then find the value of: $\frac{1+{cosec} \theta}{1-\cos \theta}$.
Answer & explanation
Correct answer: option 3
We are given that ,
sin²θ = 2sinθ - 1
sin²θ - 2sinθ + 1 = 0
sin²θ - sinθ - sinθ + 1 = 0
sinθ ( sinθ - 1 ) - 1 ( sinθ - 1 ) = 0
( sinθ - 1 )² = 0
sinθ = 1
{ using , sin90º = 1 }
So, θ = 90º
Now,
\(\frac{1 + cosecθ}{1 - cosθ}\)
= \(\frac{1 + cosec 90º}{1 - cos 90º}\)
= \(\frac{1 + 1}{1 - 0}\)
= 2