# How do you prove cot^-1(x)=pi/2-tan^-1(x)?

Kindly see the solution for the answer

#### Explanation:

The solution

Let $\alpha$ be the angle with tangent$= x$
Let $\beta = \frac{\pi}{2} - \alpha$ be the angle with cotangent $= x$

Then $\alpha + \beta = \frac{\pi}{2}$

${\cot}^{-} 1 x = \frac{\pi}{2} - {\tan}^{-} 1 x$

${\cot}^{-} 1 x = \frac{\pi}{2} - {\tan}^{-} 1 x$

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

${\cot}^{-} 1 x = \frac{\pi}{2} - \frac{\pi}{2} + {\cot}^{-} 1 x$

${\cot}^{-} 1 x = {\cot}^{-} 1 x$

God bless....I hope the explanation is useful.