Prove the following identities :

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

Multiplying numerator and denominator of L.H.S. of above equation by $\sqrt{1 + \text{cos A}}$ :

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

By formula,

1 - cos² A = sin² A

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

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

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