Question

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

Answer 1

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

Complete step-by-step answer:
Given ${\tan ^{ - 1}}\sqrt 3 + {\cot ^{ - 1}}x = \dfrac{\pi }{2} - - - - - - - > (1)$
We know that ${\tan ^{ - 1}}\sqrt 3 = \dfrac{\pi }{3}$
On substituting the above value in equation (1), we get
$ \Rightarrow \dfrac{\pi }{3} + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$
$ \Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - \dfrac{\pi }{3}$
On taking L.C.M we can rewrite the above term as
$
   \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.