Q.If $|x| < 1$, then prove that $2\tan^{-1}x = \tan^{-1}\dfrac{2x}{1-x^2} = \sin^{-1}\dfrac{2x}{1+x^2} = \cos^{-1}\dfrac{1-x^2}{1+x^2}$.

These are the identity-level properties (Properties I–V of the chapter) that let an inverse-trig expression be simplified WITHOUT drawing a triangle or invoking a sum formula — they hold strictly within the principal value branches.

Property I — undoing the outer inverse. $f^{-1}(f(\theta))=\theta$ holds only when $\theta$ already lies in $f$'s principal domain: $\sin^{-1}(\sin\theta)=\theta$ if $\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$; $\cos^{-1}(\cos\theta)=\theta$ if $\theta\in[0,\pi]$; $\tan^{-1}(\tan\theta)=\theta$ if $\theta\in\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$; similarly for $\text{cosec}^{-1},\sec^{-1},\cot^{-1}$ on their own principal domains. If $\theta$ is OUTSIDE the principal domain, $f^{-1}(f(\theta))\ne\theta$ — instead, use periodicity/symmetry to rewrite $f(\theta)$ as $f(\theta_1)$ for some $\theta_1$ that IS inside the principal domain, then $f^{-1}(f(\theta))=\theta_1$. E.g. $\sin^{-1}\left(\sin\tfrac{5\pi}6\right)=\sin^{-1}\left(\sin\left(\pi-\tfrac{\pi}6\right)\right)=\sin^{-1}\left(\sin\tfrac{\pi}6\right)=\tfrac{\pi}6$, since $\tfrac{\pi}6\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$.

Property II — undoing the inner inverse. $f(f^{-1}(x))=x$ holds throughout $f^{-1}$'s entire domain with no extra condition: $\sin(\sin^{-1}x)=x$ for $x\in[-1,1]$; $\cos(\cos^{-1}x)=x$ for $x\in[-1,1]$; $\tan(\tan^{-1}x)=x$ for every real $x$; and likewise for the other three on their own domains. This is the direct definition of "inverse" and never needs a range check.

Property III (reciprocal identities). $\sin^{-1}\left(\tfrac1x\right)=\text{cosec}^{-1}x$ and $\cos^{-1}\left(\tfrac1x\right)=\sec^{-1}x$, both for $x\in\mathbb{R}\setminus(-1,1)$; and $\tan^{-1}\left(\tfrac1x\right)=\cot^{-1}x$ if $x>0$, but $=-\pi+\cot^{-1}x$ if $x<0$ (the sign correction is needed here because $\tan^{-1}\left(\tfrac1x\right)$ stays in $\left(-\tfrac{\pi}2,0\right)$ for $x<0$ while $\cot^{-1}x$ lands in $\left(\tfrac{\pi}2,\pi\right)$ — different branches of the same underlying angle).

Property IV (reflection identities — negating the argument). $\sin^{-1}(-x)=-\sin^{-1}x$; $\tan^{-1}(-x)=-\tan^{-1}x$; $\text{cosec}^{-1}(-x)=-\text{cosec}^{-1}x$ (all three odd); but $\cos^{-1}(-x)=\pi-\cos^{-1}x$; $\sec^{-1}(-x)=\pi-\sec^{-1}x$; $\cot^{-1}(-x)=\pi-\cot^{-1}x$ (all three pick up a $\pi-{}$, since their principal range $[0,\pi]$-type interval isn't symmetric about $0$). …