Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1.$

Solving L.H.S.,

$\Rightarrow \dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \dfrac{\text{(sin θ + cos θ)}(\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ})}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ + sin θ cos θ} \\[1em] = 1.$

Since, L.H.S. = R.H.S. hence proved that $\dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1$.