Áp dụng công thức biến đổi lượng giác:
$\sin (45^{o}+x)=\sin 45^{o}\cos x+\cos 45^{o}\sin x =\dfrac{\sqrt{2}}{2}(\cos x+\sin x)$
$\cos (45^{o}-x)=\cos 45^{o}\cos x-\sin 45^{o}\sin x=\dfrac{\sqrt{2}}{2}(\cos x-\sin x)$
$A=\dfrac{\sin (45^{o}+x)-\cos (45^{o}-x)}{\sin (45^{o}+x)+\cos (45^{o}-x)}=\dfrac{\dfrac{\sqrt{2}}{2}(\cos x+\sin x)-\dfrac{\sqrt{2}}{2}(\cos x-\sin x)}{\dfrac{\sqrt{2}}{2}(\cos x+\sin x)+\dfrac{\sqrt{2}}{2}(\cos x-\sin x)}$
$=\dfrac{\sqrt{2}\sin x}{\sqrt{2}\cos x}=\tan x$