Prove a cotangent derivative and distinguish a reciprocal from an inverse

Where both sides are defined, $\cot x=1/\tan x=(\tan x)^{-1}$. Differentiating gives $-(\tan x)^{-2}\sec^2x=-\sec^2x/\tan^2x=-1/\sin^2x=-\csc^2x$, the printed proof. This particular route initially requires both $\sin x\ne0$ and $\cos x\ne0$.

Cotangent itself is defined wherever $\sin x\ne0$, including points where $\cos x=0$ and the printed reciprocal-tangent expression is undefined. An independently authored quotient-rule proof avoids that restriction: $d(\cos x/\sin x)/dx=[-\sin^2x-\cos^2x]/\sin^2x=-\csc^2x$. Thus the claimed formula holds on the full cotangent domain.

The yellow note in both copies distinguishes the reciprocal $(\tan x)^{-1}=1/\tan x$ from the inverse function $\tan^{-1}x=\arctan x$. These are not the same function. The source points to section 3.9.1 for inverse functions; its principal domains and ranges are now included in this chapter’s inverse-trigonometric guidance.