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(D) Let $g(x) = \sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$.For $x \in [0, \frac{\pi}{2}]$,the argument $(x + \frac{\pi}{4}) \in [\frac{\pi}{4

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$3 - 2\sqrt{2}$

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(D) Let $g(x) = \sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$.For $x \in [0, \frac{\pi}{2}]$,the argument $(x + \frac{\pi}{4}) \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.Thus,$g(x) \in [1, \sqrt{2}]$.Since $f(x) = 	an^{-1}(g(x))$ is an increasing function,the range of $f(x)$ is $[	an^{-1}(1), 	an^{-1}(\sqrt{2})] = [\frac{\pi}{4}, 	an^{-1}(\sqrt{2})]$.Therefore,$m = \frac{\pi}{4}$ and $M = 	an^{-1}(\sqrt{2})$.We need to find $	an(M - m) = 	an(	an^{-1}(\sqrt{2}) - \frac{\pi}{4})$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get:$	an(M - m) = \frac{\sqrt{2} - 1}{1 + \sqrt{2}(1)} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$.Rationalizing the denominator:$\frac{\sqrt{2} - 1}{\sqrt{2} + 1} 	imes \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{2 - 2\sqrt{2} + 1}{2 - 1} = 3 - 2\sqrt{2}$.

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x) = \tan^{-1}(\sin x + \cos x)$ in the interval $[0, \frac{\pi}{2}]$. Then the value of $\tan(M - m)$ is equal to:

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