How to solve this equation? $a r c t g (x) + a r c t g (\frac{1}{x}) = \frac{π}{2}$ $arctg(x)+arctg(1/x)=pi/2$ ?

Question

10789 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-06T14:18:22

The given relation is valid for all real values of x. So it is not an equation but an identity. It can be proved as follows.

Let

${tan}^{- 1} x = θ$ $tan^-1x=theta$

$\Rightarrow x = tan θ = cot (\frac{π}{2} - θ)$ $=>x=tantheta=cot(pi/2-theta)$

$\Rightarrow {cot}^{- 1} x = \frac{π}{2} - θ$ $=>cot^-1x=pi/2-theta$

$\Rightarrow {cot}^{- 1} x = \frac{π}{2} - {tan}^{- 1} x$ $=>cot^-1x=pi/2-tan^-1x$

$\Rightarrow {tan}^{- 1} x + {cot}^{- 1} x = \frac{π}{2}$ $=>tan^-1x+cot^-1x=pi/2$

$\Rightarrow {tan}^{- 1} x + {tan}^{- 1} (\frac{1}{x}) = \frac{π}{2}$ $=>tan^-1x+tan^-1(1/x)=pi/2$

Proved

How to solve this equation? arctg(x)+arctg(1x)=π2arctg(x)+arctg(1/x)=pi/2?