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

Last updated date: 18th Sep 2024
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Hint: We recall the domain range of tan inverse function that is $\arctan \left( x \right)$or ${{\tan }^{-1}}x$. We recall the relationship ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$ and the relation ${{\cot }^{-1}}x={{\tan }^{-1}}\left( \dfrac{1}{x} \right)$ for $x>0$ and ${{\cot }^{-1}}x=\pi +{{\tan }^{-1}}\left( \dfrac{1}{x} \right)$ for $x<0$. We use these identities to find the possible solutions of $x$.

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

${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$
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{align} & \Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{x}}{1-x\cdot \dfrac{1}{x}} \right) \\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{x}}{1-1} \right) \\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{x}}{0} \right) \\ \end{align}
We see that above step argument for tangent inverse function is undefined and for $x>0$ the range restricts to $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ and we have
${{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{x}}{0} \right)=\dfrac{\pi }{2}$
Hence the solution set is $\left( 0,\infty \right)$.
We should remember other reciprocal of the argument relation in tan and cot inverse function like ${{\tan }^{-1}}\left( \dfrac{1}{x} \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( x \right)$ if $x>0$ and ${{\tan }^{-1}}\left( \dfrac{1}{x} \right)=-\dfrac{\pi }{2}-{{\tan }^{-1}}\left( x \right)$ 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 $\left( 0,\pi \right)$.