Prove the following identities :

$\dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}$

By formula,

sec² - tan² A = 1

Solving L.H.S. of the equation :

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

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

Hence, proved that $\dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}$.