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

How do you prove tan⁡−1x+tan⁡−1(1x)=π2{\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = \dfrac{\pi }{2} for x>0x > 0?

Explanation

Solution

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 π2\dfrac{\pi }{2}. We have to keep the value of xx under the constraint x>0x > 0 at every step in the solution.

Formula used:
1tan⁡θ=cot⁡θ\dfrac{1}{{\tan \theta }} = \cot \theta
tan⁡a=cot⁡(π2−a)\tan a = \cot (\dfrac{\pi }{2} - a)

Complete step by step solution:
We have to prove that tan⁡−1x+tan⁡−1(1x)=π2{\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = \dfrac{\pi }{2} for x>0x > 0.
We start with the LHS, i.e. tan⁡−1x+tan⁡−1(1x){\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}). We see that two inverse trigonometric functions are added in the LHS: tan⁡−1x{\tan ^{ - 1}}x and tan⁡−1(1x){\tan ^{ - 1}}(\dfrac{1}{x}). We have to simplify these functions.
First we try to simplify the second term, i.e. tan⁡−1(1x){\tan ^{ - 1}}(\dfrac{1}{x})
Let us assume that tan⁡−1(1x)=θ{\tan ^{ - 1}}(\dfrac{1}{x}) = \theta
Then by inversion of the function tan⁡−1{\tan ^{ - 1}} from left to right side, we get,
1x=tan⁡θ ⇒x=1tan⁡θ=cot⁡θ ⇒θ=cot⁡−1x  \dfrac{1}{x} = \tan \theta \\\ \Rightarrow x = \dfrac{1}{{\tan \theta }} = \cot \theta \\\ \Rightarrow \theta = {\cot ^{ - 1}}x \\\
Therefore, tan⁡−1(1x)=θ=cot⁡−1x{\tan ^{ - 1}}(\dfrac{1}{x}) = \theta = {\cot ^{ - 1}}x, for x>0x > 0.
Thus, the LHS becomes,
tan⁡−1x+tan⁡−1(1x)=tan⁡−1x+cot⁡−1x{\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x
Now to further simplify, we assume that tan⁡−1x=a{\tan ^{ - 1}}x = a.
Then again by inversion of the function tan⁡−1{\tan ^{ - 1}} from left to right side, we get,
x=tan⁡ax = \tan a
Using the identity tan⁡a=cot⁡(π2−a)\tan a = \cot (\dfrac{\pi }{2} - a) we get,
x=cot⁡(π2−a)x = \cot (\dfrac{\pi }{2} - a)
Taking cot⁡−1{\cot ^{ - 1}} on both sides,
cot⁡−1x=cot⁡−1(cot⁡(π2−a)) cot⁡−1x=π2−a  {\cot ^{ - 1}}x = {\cot ^{ - 1}}(\cot (\dfrac{\pi }{2} - a)) \\\ {\cot ^{ - 1}}x = \dfrac{\pi }{2} - a \\\
Putting tan⁡−1x=a{\tan ^{ - 1}}x = a and cot⁡−1x=π2−a{\cot ^{ - 1}}x = \dfrac{\pi }{2} - a in the expression tan⁡−1x+cot⁡−1x{\tan ^{ - 1}}x + {\cot ^{ - 1}}x, we get,
tan⁡−1x+cot⁡−1x=a+π2−a=π2{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = a + \dfrac{\pi }{2} - a = \dfrac{\pi }{2}
Thus, we get LHS = π2\dfrac{\pi }{2} = RHS.
Hence, we proved that tan⁡−1x+tan⁡−1(1x)=π2{\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{1}{x}) = \dfrac{\pi }{2} for x>0x > 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>0x > 0) should not be violated anywhere during the solution.