Prove that : $\dfrac{\text{1 + cos A}}{\text{1 - cos A}} = \text{(cosec A + cot A)}^2$.

Multiplying numerator and denominator of L.H.S. of the given equation, by (1 + cos A) :

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

Solving R.H.S. of the given equation :

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

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

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