Question

The range of values of x is given. Solve for x, ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)=\dfrac{\pi }{3},-1 < x < 1 $

Answer 1

Hint: In this question, we have to find the value of x. But LHS has two different inverse trigonometric functions. So, we have to convert these inverse trigonometric functions into a single form of inverse trigonometric functions. We consider \[\theta ={{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)\] and then transform \[\cot \theta \] into \[\tan \theta \] . Then, solve the given equation using \[\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}\] .

Complete step-by-step solution -
Solving the LHS part, we get
\[{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)\]………………….(1)
Here, the problem is we have inverse functions of tan and cot.
First of all, we have to convert it into a single inverse function.
Let us assume, \[\theta ={{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)\]……………..(2)
Taking cot in both LHS as well as RHS in equation(2), we get
\[\cot \theta =\dfrac{1-{{x}^{2}}}{2x}\]……………….(3)
Our target is to make this cot function into a tan function so that we can have the same inverse functions in LHS.
From equation(3), we have
\[\begin{align}
  & \cot \theta =\dfrac{1-{{x}^{2}}}{2x} \\
 & \Rightarrow \dfrac{1}{\tan \theta }=\dfrac{1-{{x}^{2}}}{2x} \\
 & \Rightarrow \tan \theta =\dfrac{2x}{1-{{x}^{2}}} \\
\end{align}\]
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\]……………….(4)
We can say that equation(2) and equation(4) are equal.
\[\theta ={{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\]………………….(5)
Using equation(5), we can write equation(1) as,
\[\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \\
 & =2{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \\
\end{align}\]
Transforming the given expression, we have to solve \[2{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=\dfrac{\pi }{3}\] .
Diving by 2 in both LHS and RHS, we get
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=\dfrac{\pi }{6}\]
Now, solving this equation and using \[\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}\] , we get
\[\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=\dfrac{\pi }{6} \\
 & \Rightarrow \left( \dfrac{2x}{1-{{x}^{2}}} \right)=\tan \dfrac{\pi }{6} \\
 & \Rightarrow \left( \dfrac{2x}{1-{{x}^{2}}} \right)=\dfrac{1}{\sqrt{3}} \\
 & \Rightarrow 2\sqrt{3}x=1-{{x}^{2}} \\
 & \Rightarrow {{x}^{2}}+2\sqrt{3}x-1=0 \\
\end{align}\]
Here, we have a quadratic equation. We can get the values of x after solving this quadratic equation.
\[\begin{align}
  & x=\dfrac{-2\sqrt{3}\pm \sqrt{12-4(-1)}}{2} \\
 & \Rightarrow x=\dfrac{-2\sqrt{3}\pm \sqrt{12+4}}{2} \\
 & \Rightarrow x=\dfrac{-2\sqrt{3}\pm \sqrt{16}}{2} \\
 & \Rightarrow x=\dfrac{-2\sqrt{3}\pm 4}{2} \\
 & \Rightarrow x= - \sqrt{3}\pm {2} \\
\end{align}\]
We have $ -1 < x < 1$ .
So, \[x=2-\sqrt{3}\] .

Note: In this question, after solving the quadratic equation, we get two values of x. One can write both values of x as an answer, which is wrong. According to the information provided in the question, we have one restriction on x that is x should lie between -1 and 1. To satisfy this information, we have to take\[x=2-\sqrt{3}\] as the value of x and ignore the other value of x.