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Let $y = \tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$.First, use half-angle identities to rewrite the numerator and denominator:$$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)$$$$1 - \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$$Substitute these back into the expression:$$\frac{\cos x}{1-\sin x} = \frac{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2}$$Cancel the common term $\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)$:$$= \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$$Divide the numerator and denominator by $\cos\frac{x}{2}$:$$= \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}$$Since $\tan\frac{\pi}{4} = 1$, we can rewrite this as:$$= \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}}$$Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:$$= \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$$Therefore:$$y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$$Since $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the angle lies within the principal range, so:$$y = \frac{\pi}{4} + \frac{x}{2}$$

Let the given expression be $E$.$$E = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right]$$Evaluate each term separately using principal values:$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$(Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\tan^{-1}(-x) = -\tan^{-1}x$)$\cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$(Since $\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}$)For the third term, first evaluate the sine function:$\sin\left(-\frac{\pi}{2}\right) = -1$So, $\tan^{-1}(-1) = -\frac{\pi}{4}$Substitute these values back into the expression:$$E = -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4}$$Find a common denominator (which is 12):$$E = \frac{-2\pi + 4\pi - 3\pi}{12}$$$$E = \frac{-\pi}{12}$$

Domain We need the argument of the arcsine to be in $[-1, 1]$:$$-1 \le x^2 - 4 \le 1$$Add 4 to all parts:$$3 \le x^2 \le 5$$Taking the square root, we get two intervals:$$\sqrt{3} \le |x| \le \sqrt{5}$$This splits into positive and negative regions:$$x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$$ Range From the inequality above, we know that as $x$ varies over the domain, the term $(x^2 - 4)$ covers the entire interval $[-1, 1]$.Therefore, $f(x)$ covers all possible values of $\sin^{-1}(u)$:$$y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

Let the expression be $E$.$$E = \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)$$ Evaluate each term using principal value branches: $\tan^{-1}(1) = \frac{\pi}{4}$(Since $\tan\frac{\pi}{4} = 1$) $\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$(Since $\cos^{-1}(-x) = \pi - \cos^{-1}x$) $\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$(Since $\sin^{-1}(-x) = -\sin^{-1}x$) Substitute these values back into $E$:$$E = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{4}$$Cancel the $\frac{\pi}{4}$ terms:$$E = \frac{2\pi}{3}$$