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

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

The L.H.S of the equation can be written as,

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

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