Given that tan θ = $\dfrac{5}{12}$ and θ is an acute angle, find sin θ and cos θ.

By formula,

sec² θ = 1 + tan² θ

Substituting values we get,

$\Rightarrow \text{sec}^2 \text{ θ} = 1 + \Big(\dfrac{5}{12}\Big)^2 \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = 1 + \dfrac{25}{144} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{144 + 25}{144} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{169}{144} \\[1em] \Rightarrow \text{sec θ} = \sqrt{\dfrac{169}{144}} \\[1em] \Rightarrow \text{sec θ} = \pm \dfrac{13}{12}.$

Since, θ is an acute angle and value of sec is positive in first quadrant.

∴ sec θ = $\dfrac{13}{12}$.

By formula,

cos θ = $\dfrac{1}{\text{sec θ}} = \dfrac{1}{\dfrac{13}{12}} = \dfrac{12}{13}.$

sin² θ + cos² θ = 1

Substituting values we get,

$\Rightarrow \text{sin}^2 \text{ θ} + \Big(\dfrac{12}{13}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \Big(\dfrac{12}{13}\Big)^2\\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \dfrac{144}{169} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{169 - 144}{169} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{25}{169} \\[1em] \Rightarrow \text{sin θ} = \sqrt{\dfrac{25}{169}} \\[1em] \Rightarrow \text{sin θ} = \pm \dfrac{5}{13}.$

Since, θ is an acute angle and value of sin is positive in first quadrant.

∴ sin θ = $\dfrac{5}{13}$.

Hence, sin θ = $\dfrac{5}{13}$ and cos θ = $\dfrac{12}{13}$.