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Question

$If 3 \tan^{- 1} (\frac{1}{2 + \sqrt{3}}) - \tan^{- 1} (\frac{1}{x}) = \tan^{- 1} (\frac{1}{2})$ then 2021x is equal to

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Solution

Given equation can be written as
$\begin{array}{l} 3 \tan^{- 1} (2 - \sqrt{3}) - \tan^{- 1} (\frac{1}{x}) = \tan^{- 1} (\frac{1}{2}) \\ \Rightarrow \frac{3 π}{12} - \tan^{- 1} (\frac{1}{x}) = \tan^{- 1} (\frac{1}{2}) \\ \Rightarrow \tan^{- 1} (\frac{1}{x}) = \frac{π}{4} - \tan^{- 1} (\frac{1}{2}) \end{array}$
Taking tangent on both sides, we get
$\frac{1}{x} = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} \Rightarrow x = 3$

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