If $A={125}^{\circ },\text{ then }2\cos \frac{A}{2}$ is equal to.

$\because A={125}^{\circ }\Rightarrow \frac{A}{2}=62{\frac{1}{2}}^{\circ }$

$$\begin{gathered} \sqrt{1+\sin A}=\pm \left(\sin \frac{A}{2}+\cos \frac{A}{2}\right) \\ =\sin \frac{A}{2}+\cos \frac{A}{2} \end{gathered}$$

$\begin{array}{r}\sqrt{1-\sin A}=\pm \left(\sin \frac{A}{2}-\cos \frac{A}{2}\right)\\ =\sin \frac{A}{2}-\cos \frac{A}{2}\end{array}$

$\left(\because \frac{A}{2}=62{\frac{1}{2}}^{\circ },\text{ then }\cos \frac{A}{2}<\sin \frac{A}{2}\right)$

On subtracting Eq. (ii) from Eq. (i), We get 

$2\cos \frac{A}{2}=\sqrt{1+\sin A}-\sqrt{1-\sin A}$