Question

Solve the following equation for the value of x, ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)=\dfrac{2\pi }{3},x>0$.

Answer 1

Hint:In order to find the solution of this question, we will require a few of the inverse trigonometric formulas like, ${{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}$ and $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. We should also know the value of a few standard trigonometric ratios like $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$. We can solve this question with the help of these values.

Complete step-by-step answer:
In this question, we have been asked to find the value of x from the given equality, that is, ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)=\dfrac{2\pi }{3},x>0$. Now, to solve this equation, we should know that ${{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}$. So, we can write ${{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)$ as ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. Hence, we can apply that and write the given equation as follows,
${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=\dfrac{2\pi }{3}$
Which can also be written as,
$2{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=\dfrac{2\pi }{3}$
Now, we know that $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. So, we will apply that and write ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ as $2{{\tan }^{-1}}x$. Therefore, we can write the above equation as follows,
$\begin{align}
  & 2\times 2{{\tan }^{-1}}x=\dfrac{2\pi }{3} \\
 & \Rightarrow 4{{\tan }^{-1}}x=\dfrac{2\pi }{3} \\
\end{align}$
And we can further write it as,
$\begin{align}
  & {{\tan }^{-1}}x=\dfrac{2\pi }{3\times 4} \\
 & \Rightarrow {{\tan }^{-1}}x=\dfrac{\pi }{6} \\
\end{align}$
Now, we will take the tangent ratio of the equality. By doing so, we will get,
$\tan \left( {{\tan }^{-1}}x \right)=\tan \dfrac{\pi }{6}$
Now, we know that $\tan \left( {{\tan }^{-1}}x \right)=x$. So, we can write the above equation as,
$x=\tan \dfrac{\pi }{6}$
And we know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$. Therefore, we will get the value of x as,
$x=\dfrac{1}{\sqrt{3}}$
Hence, we can say that the value of x for the equation, given in the question is $\dfrac{1}{\sqrt{3}}$.

Note: While solving this question, one can think of taking tangent ratios at the step where we applied ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}x$. This is also correct, but lengthier and more complicated. Also, we have to be very focused and careful while doing the calculations to reduce the errors.