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

$\dfrac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{cos}^2 \text{ θ} - 1} = \text{sec θ}.$

Solving L.H.S.,

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

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{ cos}^2 \text{ θ}} = \text{sec θ}$.