If tan θ = $\dfrac{4}{3}$, find the value of sin θ + cos θ (both sin θ and cos θ are positive).

Given,

$\phantom{\Rightarrow} \text{tan θ} = \dfrac{4}{3} \\[1em] \Rightarrow \text{tan}^2 \text{ θ} = \dfrac{16}{9} \\[1em] \phantom{\Rightarrow} \text{sec}^2 \text{ θ} = 1 + \text{tan}^2 θ \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = 1 + \dfrac{16}{9} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{9 + 16}{9} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 \text{ θ}} = \dfrac{25}{9} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{9}{25} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\dfrac{9}{25}} \\[1em] \Rightarrow \text{cos θ} = \pm\dfrac{3}{5} \\[1em] \Rightarrow \text{cos θ} = \dfrac{3}{5} [\text{Given, cos θ is positive.}].$

By formula,

sin² θ + cos² θ = 1

Substituting values in sin² θ + cos² θ = 1 we get :

$\Rightarrow \text{sin}^2 \text{ θ} + \Big(\dfrac{3}{5}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \dfrac{9}{25} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{25 - 9}{25} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{16}{25} \\[1em] \Rightarrow \text{sin θ} = \sqrt{\dfrac{16}{25}} \\[1em] \Rightarrow \text{sin θ} = \pm\dfrac{4}{5} \\[1em] \Rightarrow \text{sin θ} = \dfrac{4}{5} \space [\text{Given, sin θ is positive.}].$

Substituting values in sin θ + cos θ we get :

sin θ + cos θ = $\dfrac{4}{5} + \dfrac{3}{5} = \dfrac{7}{5} = 1\dfrac{2}{5}.$

Hence, sin θ + cos θ = $1\dfrac{2}{5}$.