If sin θ + cosec θ = $3\dfrac{1}{3}$, find the value of sin² θ + cosec² θ.

Given,

$\phantom{\Rightarrow} \text{sin θ + cosec θ} = 3\dfrac{1}{3} \\[1em] \Rightarrow \text{sin θ + cosec θ} = \dfrac{10}{3} \\[1em]$

Squaring both sides we get,

$\Rightarrow \text{(sin θ + cosec θ)}^2 = \Big(\dfrac{10}{3}\Big)^2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{ 2 sin θ.cosec θ} = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 sin θ} \times \dfrac{1}{\text{sin θ}} = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 2 = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{100}{9} - 2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{100 - 18}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{82}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = 9\dfrac{1}{9}$

Hence, sin² θ + cosec² θ = $9\dfrac{1}{9}$.