\begin{aligned}
\tan x &\left.=\frac{\sin x}{\cos x} \text { (given }\right) \\
\tan (90+x)=& \frac{\sin (90+x)}{\cos (90+x)}
\end{aligned}
expanding by double angle.
\begin{aligned}
&\frac{\sin (90+x)}{\cos (90+x)}=\frac{\sin 90 \cos x+\cos 90 \sin x}{\cos 90 \cos x-\sin 90 \sin x} \\
&\text { with } \sin 90=1 \cdot \cos 90=0 . \\
&\text { we obtain } \\
&\frac{\cos x}{-\sin x}=-\cot x
\end{aligned}