Prove the following identities :

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

Solving L.H.S. of the equation :

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

By formula,

cosec² A - cot² A = 1

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

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

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