Prove the following identities :

cosec A + cot A = $\dfrac{1}{\text{cosec A - cot A}}$

Solving R.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{cosec A - cot A}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{\text{sin A}} - \dfrac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{1 - cos A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}}$

Multiplying numerator and denominator by (1 + cos A) we get :

$\Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}} \times \dfrac{\text{1 + cos A}}{\text{1 + cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A + sin A. cos A}}{\text{1 - cos}^2 A}$

By formula,

1 - cos² A = sin² A

$\Rightarrow \dfrac{\text{sin A + sin A. cos A}}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{sin}^2 A} + \dfrac{\text{sin A. cos A}}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A + cot A}.$

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

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