#### Answer

$$\tan\Big(x-\frac{\pi}{2}\Big)=\cot x$$
The statement is false.

#### Work Step by Step

$$\tan\Big(x-\frac{\pi}{2}\Big)=\cot x$$
Now what we've already known from the cofunction identities is
$$\tan\Big(\frac{\pi}{2}-x\Big)=\cot x$$
Yet here, the statement involves $\tan\Big(x-\frac{\pi}{2}\Big)$, not $\tan\Big(\frac{\pi}{2}-x\Big)$. Unfortunately, 2 expressions are not the same, so we need to find a way to transform $\tan\Big(x-\frac{\pi}{2}\Big)$ to the already known.
Though it can be a little bit tricky, we can do like this:
$$\tan\Big(x-\frac{\pi}{2}\Big)=\tan\Big[-\Big(\frac{\pi}{2}-x\Big)\Big]$$
And we already know from the negative-angle identities that
$$\tan(-\theta)=-\tan\theta$$
Therefore,
$$\tan\Big(x-\frac{\pi}{2}\Big)=-\tan\Big(\frac{\pi}{2}-x\Big)$$
$$\tan\Big(x-\frac{\pi}{2}\Big)=-\cot x$$
Since $\cot x\ne-\cot x$, $$\tan\Big(x-\frac{\pi}{2}\Big)\ne\cot x$$
The statement thus is false.