Question

How do you prove ${\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = \dfrac{\pi }{2}$ for $x > 0$?

Answer 1

Hint: Given equation is a trigonometric identity with inverse trigonometric functions. Inverse trigonometric functions are inverse functions of trigonometric functions, with restricted domains, used to obtain angle from trigonometric ratios. We have to simplify the LHS in such terms as to evaluate the value of LHS as $\dfrac{\pi }{2}$. We have to keep the value of $x$ under the constraint $x > 0$ at every step in the solution.

Formula used:
$\dfrac{1}{{\tan \theta }} = \cot \theta $
$\tan a = \cot (\dfrac{\pi }{2} - a)$

Complete step by step solution:
We have to prove that ${\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = \dfrac{\pi }{2}$ for $x > 0$.
We start with the LHS, i.e. ${\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x})$. We see that two inverse trigonometric functions are added in the LHS: ${\tan ^{ - 1}}x$ and ${\tan ^{ - 1}}(\dfrac{1}{x})$. We have to simplify these functions.
First we try to simplify the second term, i.e. ${\tan ^{ - 1}}(\dfrac{1}{x})$
Let us assume that ${\tan ^{ - 1}}(\dfrac{1}{x}) = \theta $
Then by inversion of the function ${\tan ^{ - 1}}$ from left to right side, we get,
 $
  \dfrac{1}{x} = \tan \theta \\
   \Rightarrow x = \dfrac{1}{{\tan \theta }} = \cot \theta \\
   \Rightarrow \theta = {\cot ^{ - 1}}x \\
$
Therefore, ${\tan ^{ - 1}}(\dfrac{1}{x}) = \theta = {\cot ^{ - 1}}x$, for $x > 0$.
Thus, the LHS becomes,
${\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x$
Now to further simplify, we assume that ${\tan ^{ - 1}}x = a$.
Then again by inversion of the function ${\tan ^{ - 1}}$ from left to right side, we get,
$x = \tan a$
Using the identity $\tan a = \cot (\dfrac{\pi }{2} - a)$ we get,
$x = \cot (\dfrac{\pi }{2} - a)$
Taking ${\cot ^{ - 1}}$ on both sides,
$
  {\cot ^{ - 1}}x = {\cot ^{ - 1}}(\cot (\dfrac{\pi }{2} - a)) \\
  {\cot ^{ - 1}}x = \dfrac{\pi }{2} - a \\
$
Putting ${\tan ^{ - 1}}x = a$ and ${\cot ^{ - 1}}x = \dfrac{\pi }{2} - a$ in the expression ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x$, we get,
${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = a + \dfrac{\pi }{2} - a = \dfrac{\pi }{2}$
Thus, we get LHS = \[\dfrac{\pi }{2}\] = RHS.
Hence, we proved that ${\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = \dfrac{\pi }{2}$ for $x > 0$.

Note: The above discussed solution may not be the only way to prove the given identity. We can prove an identity by various ways using different trigonometric properties or identities. We may get confused as to which identity to use where, but that can only be solved with more and more practice of such questions. Also, we have to take care that the constraint given in the question (here $x > 0$) should not be violated anywhere during the solution.