Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

$\dfrac{\text{tan θ}}{\text{1 - cot θ}} + \dfrac{\text{cot θ}}{\text{1 - tan θ}}$ = 1 + sec θ cosec θ

Solving,

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

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{tan θ}}{\text{1 - cot θ}} + \dfrac{\text{cot θ}}{\text{1 - tan θ}}$ = 1 + sec θ cosec θ.