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

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

Answered by john_santu last updated on 08/May/21

$$\Rightarrow {\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{\pi }{2}-{\cot }^{-1}\left(\frac{1+x}{1-x}\right)$$$$\Rightarrow \tan \left({\tan }^{-1}\left(\frac{1-x}{1+x}\right)\right)=\tan (\frac{\pi }{2}-{\cot }^{-1}\left(\frac{1+x}{1-x}\right))$$$$\Rightarrow \frac{1-x}{1+x}=\cot \left({\cot }^{-1}\left(\frac{1+x}{1-x}\right)\right)$$$$\Rightarrow \frac{1-x}{1+x}=\frac{1+x}{1-x}\Rightarrow {(1-x)}^2={(1+x)}^2$$$$\Rightarrow 1-2x+x^2=1+2x+x^2$$$$\Rightarrow x=0$$

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