# How do you prove tan^-1x+tan^-1(1/x)=pi/2 for x>0?

Jan 13, 2017

we first need to know the following relationships between $\tan \theta$ &$\cot \theta$

we have:

$\cot \theta = \frac{1}{\tan} \theta$

also

$\cot \theta = \tan \left(\frac{\pi}{2} - \theta\right)$

#### Explanation:

let $y = {\tan}^{- 1} x \implies x = \tan y$

$x = \tan y \implies \frac{1}{x} = \frac{1}{\tan} y = \cot y$

$\frac{1}{x} = \cot y = \tan \left(\frac{\pi}{2} - y\right)$

$\therefore \frac{\pi}{2} - y = {\tan}^{- 1} \left(\frac{1}{x}\right)$

but $y = {\tan}^{- 1} x$

so $\frac{\pi}{2} = {\tan}^{- 1} \left(\frac{1}{x}\right) + {\tan}^{- 1} x$

as reqd.

Jan 14, 2017

Take the addition formula for tangent:

$\tan \left(a + b\right) = \frac{\tan \left(a\right) + \tan \left(b\right)}{1 - \tan \left(a\right) \tan \left(b\right)}$

Rearranging:

$a + b = {\tan}^{-} 1 \left(\frac{\tan \left(a\right) + \tan \left(b\right)}{1 - \tan \left(a\right) \tan \left(b\right)}\right)$

Let $m = \tan \left(a\right)$ and $n = \tan \left(b\right)$. These both imply that $a = {\tan}^{-} 1 \left(m\right)$ and $b = {\tan}^{-} 1 \left(n\right)$.

Plugging in all these values into the before formula gives:

${\tan}^{-} 1 \left(m\right) + {\tan}^{-} 1 \left(n\right) = {\tan}^{-} 1 \left(\frac{m + n}{1 - m n}\right)$

This is applicable to the question, which asks about ${\tan}^{-} 1 \left(x\right) + {\tan}^{-} 1 \left(\frac{1}{x}\right)$. According to the rule we've just derived:

${\tan}^{-} 1 \left(x\right) + {\tan}^{-} 1 \left(\frac{1}{x}\right) = {\tan}^{-} 1 \left(\frac{x + \frac{1}{x}}{1 - x \left(\frac{1}{x}\right)}\right)$

${\tan}^{-} 1 \left(x\right) + {\tan}^{-} 1 \left(\frac{1}{x}\right) = {\tan}^{-} 1 \left(\frac{x + \frac{1}{x}}{1 - 1}\right)$

Since the denominator inside the inverse tangent function on the right-hand side is $0$, we see that the argument of the function is undefined.

For $x > 0$ and using the restricting the range of ${\tan}^{-} 1 \left(x\right)$ to $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, we see that tangent is undefined at $\frac{\pi}{2}$, so the inverse tangent of an undefined value is also $\frac{\pi}{2}$.

Therefore:

${\tan}^{-} 1 \left(x\right) + {\tan}^{-} 1 \left(\frac{1}{x}\right) = \frac{\pi}{2}$