If the equation sin^{-1}(x - 1) + cos^{-1}(x | vedclass

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(D) For the equation to be defined,the domains of the inverse trigonometric functions must be satisfied:$1) -1 \leq x - 1 \leq 1 \Rightarrow 0 \leq x

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$\frac{5\pi}{4}$

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(D) For the equation to be defined,the domains of the inverse trigonometric functions must be satisfied:$1) -1 \leq x - 1 \leq 1 \Rightarrow 0 \leq x \leq 2$$2) -1 \leq x - 3 \leq 1 \Rightarrow 2 \leq x \leq 4$$3) -x^2 + 2 
eq 0 \Rightarrow x^2 
eq 2 \Rightarrow x 
eq \pm\sqrt{2}$Taking the intersection of the domains from $(1)$ and $(2)$,we get $x = 2$.Substituting $x = 2$ into the original equation:$m = \sin^{-1}(2 - 1) + \cos^{-1}(2 - 3) + 	an^{-1}\left(\frac{2}{-2^2 + 2}ight)$$m = \sin^{-1}(1) + \cos^{-1}(-1) + 	an^{-1}\left(\frac{2}{-2}ight)$$m = \frac{\pi}{2} + \pi + 	an^{-1}(-1)$$m = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{6\pi - \pi}{4} = \frac{5\pi}{4}$

If the equation $\sin^{-1}(x - 1) + \cos^{-1}(x - 3) + \tan^{-1}\left(\frac{x}{-x^2 + 2}\right) = m$ holds,then the value of $m$ is

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