trigonometry - Contradictory (and undefined) results when using different conversions between sine and cosine

I came across this problem to solve for $x$:

$$\sin[\cot^{-1}(x + 1)]=\cos[\tan^{-1}x]$$

I tried to do it initially by converting the cosine term on the right to sine by using $\sin(\frac{\pi}{2}-x)=\cos x$, but for some reason this doesn't work:

$$\sin[\cot^{-1}(x + 1)]=\sin[\frac{\pi}{2}-\tan^{-1}x]$$

$$\implies\cot^{-1}(x + 1)=\frac{\pi}{2}-\tan^{-1}x$$

$$\implies\cot^{-1}(x + 1)=\cot^{-1}x$$

$$\implies x + 1=x$$

$$\implies 1=0$$

Yet, if instead I use $\sin(\frac{\pi}{2}+x)=\cos x$:

$$\sin[\cot^{-1}(x + 1)]=\sin[\frac{\pi}{2}+\tan^{-1}x]$$

$$\implies\cot^{-1}(x + 1)=\frac{\pi}{2}+\tan^{-1}x$$

$$\implies\frac{\pi}{2}-\tan^{-1}(x + 1)=\frac{\pi}{2}+\tan^{-1}x$$

$$\implies\tan^{-1}(-x - 1)=\tan^{-1}x$$

$$\implies-x-1=x$$

$$\implies x=-\frac{1}{2}$$

And this is indeed the correct answer according to the book. What I don't get is why the first method failed. The identity $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ should hold for all real numbers, so I don't think that is the problem. Moreover, if I plug $\frac 1 2$ and $-\frac 1 2$ into $\cot^{-1}x$, I get different answers, so this isn't a problem with the cancellation either. What's going wrong here?

EDIT: It appears that the problem is that since the value of $x$ is negative, the term $\frac \pi 2 - \tan^{-1}x$ becomes positive and lands outside the range where $\sin x$ is injective, and so the two inputs are different despite the fact that the output of their sine is equivalent. I guess a follow up question here is how one would figure this out without knowing that the answer is negative. It seems impossible to know which conversion to use beforehand without that knowledge.