If cosec x−cot x

(b) $\frac{3}{5}$

We have:

$\operatorname{cosec} x-\cot x=\frac{1}{2}$          ...(i)

$\Rightarrow \frac{1}{\operatorname{cosec} x-\cot x}=2$

$\Rightarrow \frac{\operatorname{cosec}^{2} x-\cot ^{2} x}{\operatorname{cosec} x-\cot x}=2$

$\Rightarrow \frac{(\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x)}{(\operatorname{cosec} x-\cot x)}=2$

$\therefore \operatorname{cosec} x+\cot x=2$            ...(ii)

Adding (1) and (2):

$2 \operatorname{cosec} x=\frac{1}{2}+2$

$\Rightarrow 2 \operatorname{cosec} x=\frac{5}{2}$

$\Rightarrow \operatorname{cosec} x=\frac{5}{4}$

$\Rightarrow \frac{1}{\sin x}=\frac{5}{4}$

$\Rightarrow \sin x=\frac{4}{5}$

Now, $0<\theta<\frac{\pi}{2}$

$\therefore \cos \theta=\sqrt{1-\sin ^{2} \theta}$

$=\sqrt{1-\left(\frac{4}{5}\right)^{2}}$

$=\frac{3}{5}$