Correct Answer: $\frac{3}{2}$

Solution : Given:
$\operatorname{cosec}\theta+\sin\theta=\frac{5}{2}$
⇒ $\frac{1}{\sin\theta}+\sin\theta=\frac{5}{2}$
⇒ $\frac{1+\sin^{2}\theta}{\sin\theta}=\frac{5}{2}$
⇒ $2\sin^{2}\theta-5\sin\theta+2=0$
⇒ $2\sin^{2}\theta-4\sin\theta-1\sin\theta+2=0$
⇒ $2\sin\theta (\sin\theta-2)-1(\sin\theta-2)=0$
⇒ $(\sin\theta-2)(2\sin\theta-1)=0$
⇒ $\sin\theta= 2,\frac{1}{2}$
$\sin\theta$'s value can't be 2.
So, $\sin\theta=\frac{1}{2}=\sin30^{\circ}$
⇒ $\theta=30^{\circ}$
$\operatorname{cosec}\theta-\sin\theta$
$=\operatorname{cosec}30^{\circ}-\sin30^{\circ}$
$= 2-\frac{1}{2}$
$=\frac{3}{2}$
Hence, the correct answer is $\frac{3}{2}$.