If k le sin^{-1}x + cos^{-1}x + tan^{-1}x le | vedclass

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(A) We know that for $x \in [-1, 1]$,the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ holds true.Substituting this into the given expression,we

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$k = 0, K = \pi$

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(A) We know that for $x \in [-1, 1]$,the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ holds true.Substituting this into the given expression,we get $\frac{\pi}{2} + 	an^{-1}x$.Since the domain of $	an^{-1}x$ is $(-\infty, \infty)$,the range of $	an^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.Therefore,$-\frac{\pi}{2} Adding $\frac{\pi}{2}$ to all parts of the inequality,we get $\frac{\pi}{2} - \frac{\pi}{2} This simplifies to $0 Comparing this with $k \le \sin^{-1}x + \cos^{-1}x + 	an^{-1}x \le K$,we find $k = 0$ and $K = \pi$.

If $k \le \sin^{-1}x + \cos^{-1}x + \tan^{-1}x \le K$,then

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