# How do you prove cos^-1 x + tan^-1 x = pi/2?

Apr 19, 2016

This may be true if you meant ${\cot}^{- 1} \left(x\right)$ and not ${\cos}^{- 1} \left(x\right)$. That is,
${\cot}^{- 1} \left(x\right) + {\tan}^{- 1} \left(x\right) = \setminus \frac{\pi}{2}$

#### Explanation:

Let, $\setminus \quad \tan \left(y\right) = x = \frac{x}{1}$

Because $\tan \left(\setminus \textrm{\angle}\right) = \setminus \frac{\textrm{o p p o s i t e}}{\setminus} \textrm{a \mathrm{dj} a c e n t}$, we have the diagram below:

The upper angle is $\setminus \frac{\pi}{2} - y$ because the sum of the angles of a triangle needs to be $\setminus \pi \setminus \quad$ (That is, 180 degrees).

Also note from the diagram that since $\cot \left(\setminus \textrm{\angle}\right) = \setminus \frac{\textrm{a \mathrm{dj} a c e n t}}{\setminus} \textrm{o p p o s i t e}$

we have that $\cot \left(\setminus \frac{\pi}{2} - y\right) = \frac{x}{1}$.

Invert the two expressions for $\tan$ and cot.

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

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

Therefore,

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

Note that this proof requires that we use the reference angles when taking inverses, the tan function is periodic so

$\tan \left(y + 2 \setminus \pi\right) = \tan \left(y\right) = x$

so we could take ${\tan}^{- 1} \left(x\right)$ and get $y + 2 \pi$ and this would make the identity not true. But if we take the smallest reference angles, the identity is true.