Question

How do you solve the equation? $\arctan \left(x\right)+\arctan \left({\displaystyle \frac{1}{x}}\right)={\displaystyle \frac{\pi }{2}}$

Answer 1

Hint: We recall the domain range of tan inverse function that is $\arctan \left(x\right)$or ${\tan }^{-1}x$. We recall the relationship ${\cot }^{-1}x+{\tan }^{-1}x={\displaystyle \frac{\pi }{2}}$ and the relation ${\cot }^{-1}x={\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)$ for $x>0$ and ${\cot }^{-1}x=\pi +{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)$ for $x<0$. We use these identities to find the possible solutions of $x$.

Complete step by step answer:
We know that inverse tangent function $\arctan \left(x\right)$ or ${\tan }^{-1}x$ has the domain as the real number set and the range as the interval$({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}})$. $$$$
We are given the following inverse tangent function in the question.
$$\begin{array}{rl}& \arctan \left(x\right)+\arctan \left({\displaystyle \frac{1}{x}}\right)={\displaystyle \frac{\pi }{2}}\\ & \Rightarrow {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)={\displaystyle \frac{\pi }{2}}\end{array}$$
We see clearly that $x\ne 0$since ${\displaystyle \frac{1}{x}}$ is well defined in the above equation. We know from reciprocal relation between tangent inverse and cotangent inverse that
$${\cot }^{-1}x=\{\begin{array}{cc}{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)& \text{if }x>0\\ \pi +{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)& \text{if }x<0\end{array}$$
Let us take the first case for $x>0$ and use the above identity we have
$$\begin{array}{rl}& \Rightarrow {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)\\ & \Rightarrow {\tan }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)\end{array}$$
We know from complementary angle relation that ${\tan }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)={\displaystyle \frac{\pi }{2}}$. So the given equation $\arctan \left(x\right)+\arctan \left({\displaystyle \frac{1}{x}}\right)={\displaystyle \frac{\pi }{2}}$ has the solution in the set. $(0,\mathrm{\infty })$. If we take second case
$$\begin{array}{rl}& \Rightarrow {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)\\ & \Rightarrow {\tan }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)-\pi \\ & \Rightarrow {\displaystyle \frac{\pi }{2}}-\pi ={\displaystyle \frac{-\pi }{2}}\end{array}$$
So all the values $x\in (-\mathrm{\infty },0)$ do not satisfy the given equation. So the only valid solution we have obtained is $(0,\mathrm{\infty })$.

Additional information:
Alternative method: We know from double angle formula that
$${\tan }^{-1}\left(a\right)+{\tan }^{-1}\left(b\right)={\tan }^{-1}\left({\displaystyle \frac{a+b}{1-ab}}\right)$$
The above identity is true when$ab<1$. We add $\pi$ in the right hand side of the above equation if $ab>1,a>0,b>0$ and add $-\pi$ when $ab>1,a<0,b<0$. So let us proceed from left hand side of given equation and use ten above identify to have;
$$\begin{array}{rl}& \Rightarrow {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)\\ & \Rightarrow {\tan }^{-1}\left({\displaystyle \frac{x+{\displaystyle \frac{1}{x}}}{1-x\cdot {\displaystyle \frac{1}{x}}}}\right)\\ & \Rightarrow {\tan }^{-1}\left({\displaystyle \frac{x+{\displaystyle \frac{1}{x}}}{1-1}}\right)\\ & \Rightarrow {\tan }^{-1}\left({\displaystyle \frac{x+{\displaystyle \frac{1}{x}}}{0}}\right)\end{array}$$
We see that above step argument for tangent inverse function is undefined and for $x>0$ the range restricts to $(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$ and we have
$${\tan }^{-1}\left({\displaystyle \frac{x+{\displaystyle \frac{1}{x}}}{0}}\right)={\displaystyle \frac{\pi }{2}}$$
 Hence the solution set is $(0,\mathrm{\infty })$.$$$$

Note:
We should remember other reciprocal of the argument relation in tan and cot inverse function like ${\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)={\displaystyle \frac{\pi }{2}}-{\tan }^{-1}\left(x\right)$ if $x>0$ and ${\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)=-{\displaystyle \frac{\pi }{2}}-{\tan }^{-1}\left(x\right)$ if $x<0$ for future problems. We note that just like an inverse ${\cot }^{-1}x$ has the domain the real number but its range is $(0,\pi )$.