<s>[INST] Is $\cot x = \tan (π/2 - x) $ true for any angle $x$? Is $$\cot x = \tan \Big(\frac{π}{2} - x\Big)$$ true even when $x$ is not an acute angle ?
 [/INST] True for all $x$ for which $\cot(x)$ is defined since
$\tan(\frac{\pi}{2}-x)=\frac{\sin(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)}=\frac{  \sin (\frac{\pi}{2}) \cos (x) -\cos( \frac{\pi}{2}) \sin (x)}{ \cos( \frac{\pi}{2}) \cos (x)+\sin (\frac{\pi}{2}) \sin (x) } =\frac{ \cos x }{ \sin x}= \cot (x).$
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