If we have a inverse trigonometric equation as ${{\tan }^{-1}}\left( \dfrac{x-2}{x-4} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+4} \right)=\dfrac{\pi }{4}$, then find the value of x.

Hint: Use the formula: ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ to convert the two given tan inverse functions into a single tan inverse function. The next step is to take tangent function on both the sides of the equation and use the formula: \[\tan ({{\tan }^{-1}}x)=x\], on the L.H.S. On the R.H.S use, \[\tan \dfrac{\pi }{4}=1\]. Now, solve the obtained quadratic equation to get the value of ‘x’.

Complete step-by-step solution -
We have been given: ${{\tan }^{-1}}\left( \dfrac{x-2}{x-4} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+4} \right)=\dfrac{\pi }{4}$
Applying the formula: ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$, we get,
${{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{x-2}{x-4} \right)+\left( \dfrac{x+2}{x+4} \right)}{1-\left( \dfrac{x-2}{x-4} \right)\left( \dfrac{x+2}{x+4} \right)} \right]=\dfrac{\pi }{4}$
Taking L.C.M in the numerator, we get,
${{\tan }^{-1}}\left[ \dfrac{\dfrac{\left( x-2 \right)\left( x+4 \right)+\left( x+2 \right)\left( x-4 \right)}{\left( x-4 \right)\left( x+4 \right)}}{1-\dfrac{\left( x-2 \right)\left( x+2 \right)}{\left( x-4 \right)\left( x+4 \right)}} \right]=\dfrac{\pi }{4}$
Now, using the algebraic identity: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, we get,
\[{{\tan }^{-1}}\left[ \dfrac{\dfrac{{{x}^{2}}+4x-2x-8+{{x}^{2}}-4x+2x-8}{{{x}^{2}}-16}}{1-\dfrac{{{x}^{2}}-4}{{{x}^{2}}-16}} \right]=\dfrac{\pi }{4}\]
Taking L.C.M in the denominator, we get,
\[\begin{align}
  & {{\tan }^{-1}}\left[ \dfrac{\dfrac{{{x}^{2}}+4x-2x-8+{{x}^{2}}-4x+2x-8}{{{x}^{2}}-16}}{\dfrac{{{x}^{2}}-16-\left( {{x}^{2}}-4 \right)}{{{x}^{2}}-16}} \right]=\dfrac{\pi }{4} \\
 & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{\dfrac{2{{x}^{2}}-16}{{{x}^{2}}-16}}{\dfrac{{{x}^{2}}-16-{{x}^{2}}+4}{{{x}^{2}}-16}} \right]=\dfrac{\pi }{4} \\
 & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{\dfrac{2{{x}^{2}}-16}{{{x}^{2}}-16}}{\dfrac{-12}{{{x}^{2}}-16}} \right]=\dfrac{\pi }{4} \\
\end{align}\]
Cancelling the common terms, we get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{2{{x}^{2}}-16}{-12} \right]=\dfrac{\pi }{4}\]
Taking tangent function both the sides, we get,
\[\Rightarrow \tan \left[ {{\tan }^{-1}}\left[ \dfrac{2{{x}^{2}}-16}{-12} \right] \right]=\tan \dfrac{\pi }{4}\]
Applying the formula, \[\tan ({{\tan }^{-1}}x)=x\] and using the value, \[\tan \dfrac{\pi }{4}=1\], we get,
\[\dfrac{2{{x}^{2}}-16}{-12}=1\]
By cross-multiplication, we get,
\[\begin{align}
  & 2{{x}^{2}}-16=-12 \\
 & \Rightarrow 2{{x}^{2}}=16-12 \\
 & \Rightarrow 2{{x}^{2}}=4 \\
 & \Rightarrow {{x}^{2}}=2 \\
\end{align}\]
Taking square root both sides, we get,
$x=\pm \sqrt{2}$

Note: One may note that we cannot remove the two tan inverse functions directly by taking tangent function both sides at the first step. That is a wrong approach. First, we must convert it to a single tan inverse function and then take the tangent function to both sides. Never try to convert the given tan inverse function into sine or cosine inverse function because it will make the equation much complicated.