tan-1-1-x-1-x-cot-1-1-x-1-x-pi-2-x-

Question Number 140494 by benjo_mathlover last updated on 08/May/21

\tan^{- 1} (\frac{1 - x}{1 + x}) + \cot^{- 1} (\frac{1 + x}{1 - x}) = \frac{π}{2}

x = ?

Answered by john_santu last updated on 08/May/21

\Rightarrow \tan^{- 1} (\frac{1 - x}{1 + x}) = \frac{π}{2} - \cot^{- 1} (\frac{1 + x}{1 - x})

\Rightarrow \tan (\tan^{- 1} (\frac{1 - x}{1 + x})) = \tan (\frac{π}{2} - \cot^{- 1} (\frac{1 + x}{1 - x}))

\Rightarrow \frac{1 - x}{1 + x} = \cot (\cot^{- 1} (\frac{1 + x}{1 - x}))

\Rightarrow \frac{1 - x}{1 + x} = \frac{1 + x}{1 - x} \Rightarrow {(1 - x)}^{2} = {(1 + x)}^{2}

\Rightarrow 1 - 2 x + x^{2} = 1 + 2 x + x^{2}

\Rightarrow x = 0