Recall that
[tex]\cos^2\dfrac x2=\dfrac{1+\cos x}2[/tex]
With [tex]0^\circ<x<90^\circ[/tex], we have [tex]0^\circ<\dfrac x2<45^\circ[/tex], so [tex]\cos\dfrac x2>0[/tex] and
[tex]\cos\dfrac x2=\sqrt{\dfrac{1+\cos x}2}[/tex]
Also recall that
[tex]\sec^2x=1+\tan^2x\implies\sec^2x=2\implies\sec x=\sqrt2\implies\cos x=\dfrac1{\sqrt2}[/tex]
Then
[tex]\cos\dfrac x2=\sqrt{\dfrac{1+\frac1{\sqrt2}}2}=\dfrac{\sqrt{2+\sqrt2}}2[/tex]
