Find the limits without L'Hôpital:$lim_{ x to 0 }frac{x-sin x}{x-tan x}=? $

Find the limits without L'Hôpital's rule
$$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $$
My Try:
$$\lim_{ x \to 0 }\frac{\sin(\pi-x)-\sin x}{\tan(\pi+x)-\tan x}=?\\\lim_{ x \to 0 }\frac{2\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2})}{\frac{\sin(\frac{π}{2}-x)}{\cos(\pi+x)\cos(x)}}=\lim_{ x \to 0 }\frac{(2\cos x)(-\cos x)(\cos(\frac{\pi}{2}))}{\cos x}=0$$
but:
$$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=-1/2$$
Where is my mistake?