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

& \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{aligned}$$ $$\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{aligned} & {{\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{aligned}$$ Transforming the given expression, we have to solve $$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{aligned} & \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{aligned}$$ Here, we have a quadratic equation. We can get the values of x after solving this quadratic equation. $$\begin{aligned} & 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{aligned}$$ We have $ -1 < x < 1$ . So, $$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.