$\dfrac{1}{\text{1 + cos (90° - A)}} + \dfrac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec² (90° - A)

By formula,

cos (90° - A) = sin A

Solving L.H.S. of above equation,

$\Rightarrow \dfrac{1}{\text{1 + sin A}} + \dfrac{1}{1 - \text{sin A}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A + 1 + sin A}}{\text{(1 + sin A)(1 - sin A)}} \\[1em] \Rightarrow \dfrac{2}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{cos}^2 A} \\[1em] \Rightarrow 2\text{ sec}^2 A \\[1em] \Rightarrow 2\text{ cosec}^2 (90° - A) \space [\because \text{ sec A} = \text{ cosec(90° - A)}]$

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

Hence, proved that $\dfrac{1}{\text{1 + cos (90° - A)}} + \dfrac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec² (90° - A).