Prove that $tan^{-1}x + cot^{-1}x$ = $\dfrac{\pi}{2}$

Let $tan^{-1}x = y$ ----(1)

∴ $x = tany$ ⇒ $x = cot \left( \dfrac{\pi}{2} -y \right) $

∴ $cot^{-1}x$ = $ \dfrac{\pi}{2} -y $ ----(2)


However, $y = tan^{-1}x$ from (1)

Putting this value of $y$ is equation (2), we get,

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

⇒ $tan^{-1}x + cot^{-1}x$ = $ \dfrac{\pi}{2} $

Thus proved.