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

Solution :
$$\eqalign{ & {\sin ^{ - 1}}\left( {x - 1} \right) \cr & \Rightarrow - 1 \leqslant x - 1 \leqslant 1 \cr & \Rightarrow 0 \leqslant x \leqslant 2 \cr & {\cos ^{ - 1}}\left( {x - 3} \right) \cr & \Rightarrow - 1 \leqslant x - 3 \leqslant 1 \cr & \Rightarrow 2 \leqslant x \leqslant 4 \cr & \therefore x = 2 \cr & {\text{So, }}{\sin ^{ - 1}}\left( {2 - 1} \right) + {\cos ^{ - 1}}\left( {2 - 3} \right) + {\tan ^{ - 1}}\frac{2}{{2 - 4}} = {\cos ^{ - 1}}k + \pi \cr & {\text{or, }}{\sin ^{ - 1}}1 + {\cos ^{ - 1}}\left( { - 1} \right) + {\tan ^{ - 1}}\left( { - 1} \right) = {\cos ^{ - 1}}k + \pi \cr & \Rightarrow \frac{\pi }{2} + \pi - \frac{\pi }{4} = {\cos ^{ - 1}}k + \pi \cr & \Rightarrow {\cos ^{ - 1}}k = \frac{\pi }{4}{\text{ or }}k = \frac{1}{{\sqrt 2 }} \cr} $$