Question

Let for the value of $x\in \left( 0,\dfrac{3}{2} \right)$, we have the functions as $f\left( x \right)=\sqrt{x}$, $g\left( x \right)=\tan x$ and $h\left( x \right)=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$. If the function as $\phi \left( x \right)=\left( \left( h\circ f \right)\circ g \right)\left( x \right)$, then determine the value of $\phi \left( \dfrac{\pi }{3} \right)$.
(a) $\tan \dfrac{\pi }{12}$
(b) $\tan \dfrac{7\pi }{12}$
(c) $\tan \dfrac{11\pi }{12}$
(d) $\tan \dfrac{5\pi }{12}$

Answer 1

In this question, in order to determine the value of $\phi \left( \dfrac{\pi }{3} \right)$ given that $f\left( x \right)=\sqrt{x}$, $g\left( x \right)=\tan x$ and $h\left( x \right)=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$. If $\phi \left( x \right)=\left( \left( h\circ f \right)\circ g \right)\left( x \right)$, then determine the value of $\phi \left( \dfrac{\pi }{3} \right)$ where the function $\phi$ is defined by composition of functions $f\left( x \right),g\left( x \right)$ and $h\left( x \right)$. Now in this question we will use the following trigonometric identity that $\tan \left( b-a \right)=\dfrac{\tan b-\tan a}{1+\tan a\tan b}$ and we will also be using the value $\tan \dfrac{\pi }{4}=1$ in order to determine the value of $\phi \left( \dfrac{\pi }{3} \right)$.

Complete step-by-step solution:
We are given that $x\in \left( 0,\dfrac{3}{2} \right)$.
Let the function $f$ is defined by $f\left( x \right)=\sqrt{x}$.
Let the function $g$ is defined by $g\left( x \right)=\tan x$.
Let the function $h$ is defined by $h\left( x \right)=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$
Now let us suppose the function $\phi$ is defined by composition of functions $f\left( x \right),g\left( x \right)$ and $h\left( x \right)$ given by $\phi \left( x \right)=\left( \left( h\circ f \right)\circ g \right)\left( x \right)$.
Now since we know that the composition of functions is associative.
Therefore we have

& \phi \left( x \right)=\left( \left( h\circ f \right)\circ g \right)\left( x \right) \\\ & =\left( h\circ \left( f\circ g \right) \right)\left( x \right) \end{aligned}$$ Now in order to find the function $$\phi $$, we will first determine the composition of the function $$f\circ g$$. Given that $$f\left( x \right)=\sqrt{x}$$ and $$g\left( x \right)=\tan x$$, the composition $$f\circ g$$ is given by $$\begin{aligned} & f\circ g=f\left( \tan x \right) \\\ & =\sqrt{\tan x} \end{aligned}$$ Now we will determine the composition function $$\left( h\circ \left( f\circ g \right) \right)\left( x \right)$$ using $$f\circ g=\sqrt{\tan x}$$ and $$h\left( x \right)=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$$. We have $$\begin{aligned} & \left( h\circ \left( f\circ g \right) \right)\left( x \right)=h\left( f\circ g \right) \\\ & =h\left( \sqrt{\tan x} \right) \\\ & =\dfrac{1-{{\left( \sqrt{\tan x} \right)}^{2}}}{1+{{\left( \sqrt{\tan x} \right)}^{2}}} \\\ & =\dfrac{1-\tan x}{1+\tan x} \\\ & =\dfrac{1+\left( -\tan x \right)}{1-\left( -\tan x \right)} \end{aligned}$$ Now since we know that $$\tan \dfrac{\pi }{4}=1$$ and since $$\tan \left( -\dfrac{\pi }{4} \right)=-1$$, using this we get that $$\begin{aligned} & \left( h\circ \left( f\circ g \right) \right)\left( x \right)=\dfrac{1+\left( -\tan x \right)}{1-\left( -\tan x \right)} \\\ & =-\left( \dfrac{\tan \dfrac{\pi }{4}+\left( -\tan x \right)}{1-\left( \tan \left( \dfrac{\pi }{4} \right)\left( -\tan x \right) \right)} \right) \\\ & =\left( \dfrac{\tan \dfrac{\pi }{4}+\tan \left( -x \right)}{1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)} \right) \end{aligned}$$ Where $$\begin{aligned} & 1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)=1-\tan \left( -x \right)\left( 1 \right) \\\ & =1+\tan x \end{aligned}$$ Now since we know the following trigonometric identity that $$\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$$. On comparing the expression $$\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$$ with the expression $$\left( \dfrac{\tan \dfrac{\pi }{4}+\tan \left( -x \right)}{1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)} \right)$$, we get that $$b=-x$$ and $$a=\dfrac{\pi }{4}$$. Now we have $$\begin{aligned} & \left( h\circ \left( f\circ g \right) \right)\left( x \right)=-\left( \dfrac{\tan \dfrac{\pi }{4}+\tan \left( -x \right)}{1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)} \right) \\\ & =\tan \left( \dfrac{\pi }{4}+\left( -x \right) \right) \\\ & =\tan \left( \dfrac{\pi }{4}-x \right) \end{aligned}$$ Therefore since $$\phi $$ is defined by $$\phi \left( x \right)=\left( \left( h\circ f \right)\circ g \right)\left( x \right)$$, We have $$\phi \left( x \right)=\tan \left( \dfrac{\pi }{4}-x \right)$$ Now in order to find the value of $$\phi \left( \dfrac{\pi }{3} \right)$$, we will substitute $$x=\dfrac{\pi }{3}$$ in $$\phi \left( x \right)=\tan \left( \dfrac{\pi }{4}-x \right)$$. Then we get $$\begin{aligned} & \phi \left( \dfrac{\pi }{3} \right)=\tan \left( \dfrac{\pi }{4}-\dfrac{\pi }{3} \right) \\\ & =\tan \left( \dfrac{3\pi -4\pi }{12} \right) \\\ & =\tan \left( \dfrac{-\pi }{12} \right) \end{aligned}$$ Now since $$\tan \left( -x \right)=-\tan x$$, therefore we have $$\phi \left( \dfrac{\pi }{3} \right)=-\tan \left( \dfrac{\pi }{12} \right)$$ Also since $$-\tan x=\tan \left( \pi -x \right)$$, the above expression becomes $$\begin{aligned} & \phi \left( \dfrac{\pi }{3} \right)=-\tan \left( \dfrac{\pi }{12} \right) \\\ & =\tan \left( \pi -\dfrac{\pi }{12} \right) \\\ & =\tan \dfrac{11\pi }{12} \end{aligned}$$ Therefore we get that $$\phi \left( \dfrac{\pi }{3} \right)=\tan \dfrac{11\pi }{12}$$. **Hence option (c) is correct.** **Note:** In this problem, in order to determine the value of $$\phi \left( \dfrac{\pi }{3} \right)$$ we are using the following identities. We have $$\tan \left( b-a \right)=\dfrac{\tan b-\tan a}{1+\tan a\tan b}$$ and we will also be using the value $$\tan \dfrac{\pi }{4}=1$$ , $$\tan \left( -x \right)=-\tan x$$ and $$-\tan x=\tan \left( \pi -x \right)$$.