If $ \sin \mathrm{A}=\frac{1}{2} $, then the value of $ \cot \mathrm{A} $ is
(A) $ \sqrt{3} $
(B) $ \frac{1}{\sqrt{3}} $
(C) $ \frac{\sqrt{3}}{2} $
(D) 1

Given:

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

To do:

We have to find the value of $cot\ A$.

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $sin\ A=\frac{1}{2}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

$cot\ \theta=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow (2)^2=(AB)^2+1^2$

$\Rightarrow AB^2=4-1$

$\Rightarrow AB=\sqrt{3}$

Therefore,

$cot\ \A=\frac{AB}{BC}=\frac{\sqrt3}{1}$

$=\sqrt3$.