If sin θ = $\dfrac{\sqrt{3}}{2}$, find the value of (cosec θ + cot θ).

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} =\dfrac{\sqrt{3}}{2}$

Let perpendicular = $\sqrt{3}x$ and hypotenuse = 2x

We will find the value of base using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (2x)² - $(\sqrt{3}x)^2$

Base² = 4x² - 3x²

Base² = x²

Base = x

cosec θ = $\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{2x}{\sqrt{3}x} = \dfrac{2}{\sqrt{3}}$

cot θ = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{x}{\sqrt{3}x} = \dfrac{1}{\sqrt{3}}$

Substituting above values in cosec θ + cot θ, we get :

cosec θ + cot θ = $\dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}}$

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

Hence, cosec θ + cot θ = $\sqrt{3}$.