If $\sin ^2 \theta=2 \sin \theta-1, 0^{\circ} \leq \theta \leq 90^{\circ}$, then find the value of: $\frac{1+{cosec} \theta}{1-\cos \theta}$.

2

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