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

$\dfrac{\text{sin θ}}{\text{cot θ + cosec θ}} = 2 + \dfrac{\text{sin θ}}{\text{cot θ - cosec θ}}$

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

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

R.H.S. of the equation can be written as,

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

Since, L.H.S. = 1 - cos θ = R.H.S. hence proved that,

$\dfrac{\text{sin θ}}{\text{cot θ + cosec θ}} = 2 + \dfrac{\text{sin θ}}{\text{cot θ - cosec θ}}$.