For \(x \in \left( {0,\frac{3}{2}} \right),{\rm{\;let\;}}f\left( x \right) = \sqrt x\), g(x) = tanx and \(h\left( x \right) = \frac{{1 - {x^2}}}{{1 + {x^2}}}\). If ϕ(x) = ((hof)og)(x), then \(\phi \left( {\frac{\pi }{3}} \right)\) is equal to:

For \(x \in \left( {0,\frac{3}{2}} \right),{\rm{\;let\;}}f\left( x \ri

1.	For \(x \in \left( {0,\frac{3}{2}} \right),{\rm{\;let\;}}f\left( x \right) = \sqrt x\), g(x) = tanx and \(h\left( x \right) = \frac{{1 - {x^2}}}{{1 + {x^2}}}\). If ϕ(x) = ((hof)og)(x), then \(\phi \left( {\frac{\pi }{3}} \right)\) is equal to:
A.	\(tan\frac{\pi }{{12}}\)
B.	\(tan\frac{{11\pi }}{{12}}\)
C.	\(tan\frac{{7\pi }}{{12}}\)
D.	\(tan\frac{{5\pi }}{{12}}\)
Answer» C. \(tan\frac{{7\pi }}{{12}}\)

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For \(x \in \left( {0,\frac{3}{2}} \right),{\rm{\;let\;}}f\left( x \right) = \sqrt x\), g(x) = tanx and \(h\left( x \right) = \frac{{1 - {x^2}}}{{1 + {x^2}}}\). If ϕ(x) = ((hof)og)(x), then \(\phi \left( {\frac{\pi }{3}} \right)\) is equal to:

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