If \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to

\(\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}\)

\(\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}\)

If \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}}

1.	If \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to
A.	0
B.	1
C.	\(\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}\)
D.	\(\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}\)
Answer» B. 1

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If \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to

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