For x ∈(0,3 / 2), let f(x)=√ x, g(x)= tan x and h (x)=(1-x^2)/(1+x^2). If φ(x)=(( h ° f) ° g)(x), then φ((π)/3) is equal to:

For $\displaystyle x \in(0,3 / 2)$, let $\displaystyle f(x)=\sqrt{x}, g(x)=\tan x$ and $\displaystyle \mathrm{h}(x)=\frac{1-x^2}{1+x^2}$. If $\displaystyle \phi(x)=((\mathrm{h} \circ f) \circ g)(x)$, then $\displaystyle \phi\left(\frac{\pi}{3}\right)$ is equal to :