If sin A = $\dfrac{1}{2}$, then the value of cot A is

1. $\sqrt{3}$

2. $\dfrac{1}{\sqrt{3}}$

3. $\dfrac{\sqrt{3}}{2}$

4. 1

Let ABC be a right angle triangle with ∠B = 90°.

By formula,

sin A = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{2} = \dfrac{BC}{AC}$

Let BC = x and AC = 2x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (2x)² = AB² + (x)²

⇒ 4x² = AB² + x²

⇒ AB² = 3x²

⇒ AB = $\sqrt{3x^2} = \sqrt{3}x$.

By formula,

cot A = $\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{AB}{BC} = \dfrac{\sqrt{3}x}{x} = \sqrt{3}$.

Hence, Option 1 is the correct option.