Q.Is $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$ true? Justify your answer.

Sine, cosine and tangent all fail to be one-to-one over their full domains, so each is first restricted to an interval on which it is a bijection, and the inverse is defined only on that restricted piece.

Inverse sine. Restrict $\sin x$ to $\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$ (the interval on which it is one-to-one AND onto $[-1,1]$). Then $\sin^{-1}:[-1,1]\to\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$ is defined by $\sin^{-1}x=y \iff \sin y = x$ and $y\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$. Its graph is the mirror image of the restricted sine curve in the line $y=x$: an increasing S-shaped curve through the origin, from $\left(-1,-\tfrac{\pi}2\right)$ to $\left(1,\tfrac{\pi}2\right)$. Since sine is odd on its restricted domain, $\sin^{-1}x$ is also odd: $\sin^{-1}(-x)=-\sin^{-1}x$.

Inverse cosine. Restrict $\cos x$ to $[0,\pi]$ (one-to-one, onto $[-1,1]$). Then $\cos^{-1}:[-1,1]\to[0,\pi]$ is defined by $\cos^{-1}x=y\iff\cos y=x$ and $y\in[0,\pi]$. Its graph falls from $(-1,\pi)$ through $\left(0,\tfrac{\pi}2\right)$ to $(1,0)$. Cosine on $[0,\pi]$ is neither even nor odd (evenness would need a domain symmetric about $0$ where it stays one-to-one, which is impossible for a periodic function away from the origin), so $\cos^{-1}x$ is neither even nor odd either; a useful substitute identity is $\cos^{-1}(-x)=\pi-\cos^{-1}x$.

Inverse tangent. Restrict $\tan x$ to $\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$ (one-to-one, onto all of $\mathbb{R}$ since tangent has vertical asymptotes exactly at the endpoints). Then $\tan^{-1}:\mathbb{R}\to\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$ is defined by $\tan^{-1}x=y\iff\tan y=x$ and $y\in\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$. Its graph is a flattening S-curve through the origin that hugs the two horizontal asymptotes $y=\pm\tfrac{\pi}2$ without ever touching them, since $\tan^{-1}x$ is defined for every real $x$ (unlike $\sin^{-1},\cos^{-1}$, whose domain stops at $\pm1$). Tangent is odd on its restricted domain, so $\tan^{-1}x$ is odd.

Working with a domain question. Because $\sin^{-1}$ and $\cos^{-1}$ only accept inputs in $[-1,1]$, a composite expression like $\sin^{-1}(g(x))$ has domain exactly $\{x: -1\le g(x)\le1\}$, solved as a compound inequality. $\tan^{-1}(g(x))$, by contrast, places no restriction on $g(x)$ itself (any real value is accepted) — the only restriction comes from whatever makes $g(x)$ itself well defined (e.g. a square root needing a non-negative radicand).