If sin θ = $\dfrac{6}{10}$, find the value of cos θ + tan θ.

Given,

$\Rightarrow \text{sin θ} = \dfrac{6}{10} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{36}{100} \\[1em] \Rightarrow 1 - \text{cos}^2 \text{ θ} = \dfrac{36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = 1 - \dfrac{36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{100 - 36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{64}{100} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\dfrac{64}{100}} \\[1em] \Rightarrow \text{cos θ} = \dfrac{8}{10} \\[1.5em] \text{tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}} \\[1em] = \dfrac{\dfrac{6}{10}}{\dfrac{8}{10}} = \dfrac{6}{8}.$

Substituting values in cos θ + tan θ we get,

$\text{cos θ + tan θ }= \dfrac{8}{10} + \dfrac{6}{8} \\[1em] = \dfrac{32 + 30}{40} \\[1em] = \dfrac{62}{40} \\[1em] = \dfrac{31}{20} \\[1em] = 1\dfrac{11}{20}$

Hence, cos θ + tan θ = $1\dfrac{11}{20}.$