In the figure (2) given below, △ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin² θ + tan² θ.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (2)² + (1)²

⇒ AC² = 4 + 1 = 5

⇒ AC = $\sqrt{5}$.

Calculating sin θ, we get :

$\text{sin θ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} \\[1em] = \dfrac{2}{\sqrt{5}}.$

Calculating tan θ, we get :

$\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} \\[1em] = \dfrac{2}{1} = 2.$

Substituting value of sin θ and tan θ in sin² θ + tan² θ, we get :

$\text{sin}^2 \text{ θ} + \text{tan}^2 \text{ θ} = \Big(\dfrac{2}{\sqrt{5}}\Big)^2 + 2^2 \\[1em] = \dfrac{4}{5} + 4 \\[1em] = \dfrac{4 + 20}{5} \\[1em] = \dfrac{24}{5} = 4\dfrac{4}{5}.$

Hence, sin² θ + tan² θ = $4\dfrac{4}{5}$.