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

Hint: 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
\[\begin{align}
  & \phi \left( x \right)=\left( \left( h\circ f \right)\circ g \right)\left( x \right) \\
 & =\left( h\circ \left( f\circ g \right) \right)\left( x \right)
\end{align}\]
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
\[\begin{align}
  & f\circ g=f\left( \tan x \right) \\
 & =\sqrt{\tan x}
\end{align}\]
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
\[\begin{align}
  & \left( h\circ \left( f\circ g \right) \right)\left( x \right)=h\left( f\circ g \right) \\
 & =h\left( \sqrt{\tan x} \right) \\
 & =\dfrac{1-{{\left( \sqrt{\tan x} \right)}^{2}}}{1+{{\left( \sqrt{\tan x} \right)}^{2}}} \\
 & =\dfrac{1-\tan x}{1+\tan x} \\
 & =\dfrac{1+\left( -\tan x \right)}{1-\left( -\tan x \right)}
\end{align}\]
Now since we know that \[\tan \dfrac{\pi }{4}=1\] and since \[\tan \left( -\dfrac{\pi }{4} \right)=-1\], using this we get that
\[\begin{align}
  & \left( h\circ \left( f\circ g \right) \right)\left( x \right)=\dfrac{1+\left( -\tan x \right)}{1-\left( -\tan x \right)} \\
 & =-\left( \dfrac{\tan \dfrac{\pi }{4}+\left( -\tan x \right)}{1-\left( \tan \left( \dfrac{\pi }{4} \right)\left( -\tan x \right) \right)} \right) \\
 & =\left( \dfrac{\tan \dfrac{\pi }{4}+\tan \left( -x \right)}{1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)} \right)
\end{align}\]
Where
 \[\begin{align}
  & 1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)=1-\tan \left( -x \right)\left( 1 \right) \\
 & =1+\tan x
\end{align}\]
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
\[\begin{align}
  & \left( h\circ \left( f\circ g \right) \right)\left( x \right)=-\left( \dfrac{\tan \dfrac{\pi }{4}+\tan \left( -x \right)}{1-\tan \left( \dfrac{\pi }{4} \right)\tan \left( -x \right)} \right) \\
 & =\tan \left( \dfrac{\pi }{4}+\left( -x \right) \right) \\
 & =\tan \left( \dfrac{\pi }{4}-x \right)
\end{align}\]
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
\[\begin{align}
  & \phi \left( \dfrac{\pi }{3} \right)=\tan \left( \dfrac{\pi }{4}-\dfrac{\pi }{3} \right) \\
 & =\tan \left( \dfrac{3\pi -4\pi }{12} \right) \\
 & =\tan \left( \dfrac{-\pi }{12} \right)
\end{align}\]
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
\[\begin{align}
  & \phi \left( \dfrac{\pi }{3} \right)=-\tan \left( \dfrac{\pi }{12} \right) \\
 & =\tan \left( \pi -\dfrac{\pi }{12} \right) \\
 & =\tan \dfrac{11\pi }{12}
\end{align}\]
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 \[-\tan x=\tan \left( \pi -x \right)\].