Correct Answer - Option 1 : 0
Concept:
- \({\sec ^{ - 1}}{\rm{A}} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{1}{{\rm{A}}}} \right),\)
- \({\sin ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2} - {\cos ^{ - 1}}{\rm{x}} \Rightarrow {\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2}\)
Calculation:
Given: \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\) ---(1)
We know, \({\cos ^{ - 1}}{\rm{x}} = {\sec ^1}\left( {\frac{1}{{\rm{x}}}} \right)\)
\(\therefore {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) = {\cos ^{ - 1}}\left( {\frac{1}{{\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}}}} \right)\)
\(= {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)
\({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)
\(\Rightarrow {\rm{y}} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)
\(\Rightarrow {\rm{y}} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\) \(\left( {{{\sin }^{ - 1}}{\rm{x}} + {{\cos }^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2}} \right)\)
\(\Rightarrow {\rm{y}} = \frac{{\rm{\pi }}}{2}\)
\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 0\)
Hence, option (1) is correct.
