$sin\theta =0.5$, then  $cos\theta .tan\theta .sec\theta .csc\theta .cot\theta =$

$cos\theta .tan\theta .sec\theta .csc\theta .cot\theta =(cos\theta \times sec\theta )(tan\theta \times cot\theta )csc\theta$

$=\frac{1}{\sin \theta }=\frac{1}{0.5}=2$

                                            (OR)

Given $\sin \theta =0.5$

$\sin \theta =\frac{1}{2}$

$=\frac{sideoppto\theta }{Hypotenuse}=\frac{1}{2}$

$\frac{PQ}{PR}=\frac{1}{2}$

In right angled triangle $PQR,\overline{)Q}=90°$

$P{R}^2=P{Q}^2+Q{R}^2$

${\left(2\right)}^2={\left(1\right)}^2+Q{R}^2$

$4-1=Q{R}^2$

$3=Q{R}^2$

$QR=\sqrt{3}$

$\Delta PQR$

$\cos \theta =\frac{\sqrt{3}}{2},\tan \theta =\frac{1}{\sqrt{3}},\sec \theta =\frac{2}{\sqrt{3}},\cos ec\theta =\frac{2}{1}$$and\cot \theta =\sqrt{3}$

$\cos \theta .\tan \theta .\sec \theta .\cos ec\theta .\cot \theta$

$=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}\times \frac{2}{\sqrt{3}}\times 2\times \sqrt{3}$

$=2$