Question
यदि $k \le {\sin ^{ - 1}}x + {\cos ^{ - 1}}x + {\tan ^{ - 1}}x \le K,$ तब
${\sin ^{ - 1}}x + {\cos ^{ - 1}}x + {\tan ^{ - 1}}x = \frac{\pi }{2} + {\tan ^{ - 1}}x$
Since $\frac{{ - \pi }}{2} \le {\tan ^{ - 1}}x \le \frac{\pi }{2} $
$\Rightarrow 0 \le \frac{\pi }{2} + {\tan ^{ - 1}}x \le \pi $
$\therefore$ $K = \pi ,k = 0$.
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