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

$\dfrac{\text{cot θ}}{\text{cosec θ + 1}} + \dfrac{\text{cosec θ + 1}}{\text{cot θ}} = \text{2 sec θ}.$

Solving L.H.S.,

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

Since, L.H.S. = R.H.S hence proved that $\dfrac{\text{cot θ}}{\text{cosec θ + 1}} + \dfrac{\text{cosec θ + 1}}{\text{cot θ}} = 2\text{ sec θ}$.