For $x \in\left(0, \frac{3}{2}\right)$, let $f(x)=\sqrt{x}, g(x)=\tan x$ and
$h(x)=\frac{1-x^{2}}{1+x^{2}} .$ If $\phi(x)=(($ hof $) \circ g)(x)$, then
$\phi=\left(\frac{\pi}{3}\right)$ is equal to :
Correct Option: , 3
$f(x)=\sqrt{x}, g(x)=\tan x, h(x)=\frac{1-x^{2}}{1+x^{2}}$
$f o g(x)=\sqrt{\tan x}$
$\operatorname{hofog}(x)=h(\sqrt{\tan x})=\frac{1-\tan x}{1+\tan x}$
$=-\tan \left(\frac{\pi}{4}-x\right)$
$\phi(x)=\tan \left(\frac{\pi}{4}-x\right)$
$\phi\left(\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{4}-\frac{\pi}{3}\right)=\tan \left(-\frac{\pi}{12}\right)=-\tan \frac{\pi}{12}$
$=\tan \left(\pi-\frac{\pi}{12}\right)=\tan \frac{11 \pi}{12}$
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