A) \[k=0,\,K=\pi \]
B) \[k=0,K=\frac{\pi }{2}\]
C) \[k=\frac{\pi }{2},K=\pi \]
D) None of these
Correct Answer: A
Solution :
We have \[{{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x=\frac{\pi }{2}+{{\tan }^{-1}}x\] Since \[\frac{-\pi }{2}\le {{\tan }^{-1}}x\le \frac{\pi }{2}\Rightarrow 0\le \frac{\pi }{2}+{{\tan }^{-1}}x\le \pi \] \ \[K=\pi ,k=0\].
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