Question

If ${\tan ^{ - 1}}\sqrt 3 + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ then find the value of x.

Hint: To solve this problem we need to have basic knowledge about trigonometric and its values, and also basic calculation for simplifying the values.

Given ${\tan ^{ - 1}}\sqrt 3 + {\cot ^{ - 1}}x = \dfrac{\pi }{2} - - - - - - - > (1)$
We know that ${\tan ^{ - 1}}\sqrt 3 = \dfrac{\pi }{3}$
$\Rightarrow \dfrac{\pi }{3} + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$
$\Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - \dfrac{\pi }{3}$
$\Rightarrow {\cot ^{ - 1}}x = \dfrac{{3\pi - 2\pi }}{6} \\ \Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{6} \\$
$\Rightarrow x = \cot \dfrac{\pi }{6} = \sqrt 3 \\ \Rightarrow x = \sqrt 3 \\$
Therefore the value of $x = \sqrt 3$.
Note: In this problem we took the ${\tan ^{ - 1}}\sqrt 3$ value and replaced it in the equation. Later we have simplified the $\pi$ values which is equal to cot inverse x and on further simplification we got x value. In this kind of problem instead of solving the equation it’s better to find the value of the term and simplify.