For x in left( 0, frac{3}{2} right),let f(x) | vedclass

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(A) Given functions are $f(x) = \sqrt{x}$,$g(x) = 	an x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$.First,find $(f \circ g)(x) = f(g(x)) = \sqrt{	an x}$.N

Vedclass · Accepted Answer

$	an \frac{11\pi}{12}$

Vedclass · Answer

(A) Given functions are $f(x) = \sqrt{x}$,$g(x) = 	an x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$.First,find $(f \circ g)(x) = f(g(x)) = \sqrt{	an x}$.Next,find $\phi(x) = (h \circ (f \circ g))(x) = h(\sqrt{	an x})$.Substituting $\sqrt{	an x}$ into $h(x)$,we get $\phi(x) = \frac{1 - (\sqrt{	an x})^2}{1 + (\sqrt{	an x})^2} = \frac{1 - 	an x}{1 + 	an x}$.Using the trigonometric identity $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we can write $\frac{1 - 	an x}{1 + 	an x} = 	an\left( \frac{\pi}{4} - x ight)$.Now,calculate $\phi\left( \frac{\pi}{3} ight) = 	an\left( \frac{\pi}{4} - \frac{\pi}{3} ight) = 	an\left( \frac{3\pi - 4\pi}{12} ight) = 	an\left( -\frac{\pi}{12} ight) = -	an\left( \frac{\pi}{12} ight)$.Since $	an(\pi - 	heta) = -	an 	heta$,we have $-	an\left( \frac{\pi}{12} ight) = 	an\left( \pi - \frac{\pi}{12} ight) = 	an\left( \frac{11\pi}{12} ight)$.

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 $\phi(x) = ((h \circ f) \circ g)(x)$,then $\phi\left( \frac{\pi}{3} \right)$ is equal to

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