$F o r x \in (0, \frac{3}{2}), l e t f (x) = \sqrt{x}, g (x) = \tan x a n d h (x) = \frac{1 - x^{2}}{1 + x^{2}} . I f ϕ (x) = ((h o f) o g) (x) t h e n ϕ (\frac{π}{3}) i s e q u a l t o$

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$\tan \frac{5 π}{12}$

$\tan \frac{π}{12}$

$\tan \frac{7 π}{12}$

$\tan \frac{11 π}{12}$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{x}, g (x) = T a n x, h (x) = \frac{1 - x^{2}}{1 + x^{2}} \\ ((h o f) o g) x = (h o f) [g (x)] = (h o f) (\tan x) \\ = h [f (\tan x)] \\ = h [\sqrt{\tan x}] \\ = \frac{1 - \tan x}{1 + \tan x} \\ = T a n (\frac{5 π}{4} - x) \\ ϕ (\frac{π}{3}) = \tan (\frac{5 π}{4} - x) \\ ϕ (\frac{π}{3}) = \tan (\frac{5 π}{4} - \frac{π}{3}) \\ = \tan (\frac{11 π}{12}) \end{array}$