What is $\tan^{-1}(x) - \tan^{-1}\big(\frac{1}{x}\big)$?

Question

From the question that I'm trying to solve, going backwards it seems that the answer should be $\tan^{-1}\frac{x-\frac{1}{x}}{2}$, but I have don't know how to show this is true.

Edit: One way this question is relevant is when looking at the effect of different components of a transfer function in determining the phase of the plant at a particular frequency. It is often useful to formulate this in terms of the phase of individual sections of the transfer function of the plant and then looking at the overall effect. That's how I came across this problem as I was trying to write the expression for the phase of a system with a particular transfer function in a simplified way to draw some insights. The answers have been very helpful as I can now see that these problems in general can be simplified by using the tan double angle formula.

Answer 1

By using $$\tan{(A-B)}=\frac{\tan{(A)}-\tan{(B)}}{1+\tan{(A)}\tan{(B)}}$$ we have that $$\tan{(\arctan{(x)}-\arctan{(\frac1x)})}=\frac{x-\frac1x}{1+x(\frac1x)}=\frac{x-\frac1x}{2}$$ So the given expression is $$\arctan{(x)}-\arctan{(\frac1x)}=\arctan{(\frac{x-\frac1x}{2})}$$

Answer 2

$$\tan (\tan^{-1}(x) - \tan^{-1}(\frac{1}{x}))= {x-{1\over x}\over 1+x{1\over x}} ={x-{1\over x}\over 2}$$

So the answer is $$\tan^{-1}({x^2-1\over 2x})$$

Answer 3

Let $\theta = \tan^{-1} x$ and draw a triangle that shows this. That is, $\tan \theta = x$, so let the opposite leg be $x$ and the adjacent leg be $1$.

What angel is $\phi = \tan^{-1}(1/x)$? You have $\tan \phi = 1/x$, so $\phi$ must be the complementary angle to $\theta.$ So $$\tan^{-1} x +\tan^{-1}(1/x)=\theta+\phi = \pi/2.$$

Edit: Well, I read the problem with a plus instead of a minus. I'll leave this up for a while, because I think this is the right way to analyze such things.

Answer 4

you can get another form $$(\arctan{(x)}-\arctan{(\frac1x)})'=\frac{2}{1+x^2}$$ integrate $\frac{2}{1+x^2}$ to get $$\arctan{(x)}-\arctan{(\frac1x)}=2\arctan x-\frac{\pi}{2} ..............x>0$$

Answer 5

The function $$ f(x)=\arctan x-\arctan\frac{1}{x} $$ is defined for $x\ne0$. The derivative is $$ f'(x)=\frac{1}{1+x^2}-\frac{1}{1+\dfrac{1}{x^2}}\frac{-1}{x^2}=\frac{2}{1+x^2} $$ Hence we can say that there exists constants $c_+$ and $c_-$ such that $$ \arctan x-\arctan\frac{1}{x}= \begin{cases} c_++2\arctan x & x>0 \\[4px] c_-+2\arctan x & x<0 \end{cases} $$ Evaluating at $1$ yields $f(1)=0$, so $c_+=-\pi/2$; evaluating at $-1$ yields $f(-1)=0$, so $c_-=\pi/2$.

Thus $$ f(x)= \begin{cases} 2\arctan x-\dfrac{\pi}{2} & x>0 \\[4px] 2\arctan x+\dfrac{\pi}{2} & x<0 \end{cases} $$