If \(2{\rm{y}} = {\left( {{\rm{co}}{{\rm{t}}^{ - 1}}\left( {\frac{{\sqrt 3 {\rm{cos\;x}} + {\rm{sin\;x}}}}{{{\rm{cos\;x}} - \sqrt 3 {\rm{sin\;x}}}}} \right)} \right)^2},\;x \in \left( {0,\frac{\pi }{2}} \right)\;{\rm{then}}\;\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to

\({\rm{\;}}\frac{{\rm{\pi }}}{3} - x\)

\({\rm{\;}}\frac{{\rm{\pi }}}{6} - x\)

\({\rm{\;}}2x - \frac{{\rm{\pi }}}{3}\)

If \(2{\rm{y}} = {\left( {{\rm{co}}{{\rm{t}}^{ - 1}}\left( {\frac{{\sq

1.	If \(2{\rm{y}} = {\left( {{\rm{co}}{{\rm{t}}^{ - 1}}\left( {\frac{{\sqrt 3 {\rm{cos\;x}} + {\rm{sin\;x}}}}{{{\rm{cos\;x}} - \sqrt 3 {\rm{sin\;x}}}}} \right)} \right)^2},\;x \in \left( {0,\frac{\pi }{2}} \right)\;{\rm{then}}\;\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to
A.	\({\rm{\;}}\frac{{\rm{\pi }}}{6} - x\)
B.	\(x - \frac{{\rm{\pi }}}{6}\)
C.	\({\rm{\;}}\frac{{\rm{\pi }}}{3} - x\)
D.	\({\rm{\;}}2x - \frac{{\rm{\pi }}}{3}\)
Answer» C. \({\rm{\;}}\frac{{\rm{\pi }}}{3} - x\)

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