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Question Number 27393 by kumar123 last updated on 06/Jan/18

$$solve$$$${\tan }^{-1}(2x/1-x^2)+{\cot }^{-1}(1-x^2/2x)=\pi /3$$

Answered by $@ty@m last updated on 06/Jan/18

$$Letx=\tan u$$$${\tan }^{-1}\frac{2\tan u}{1-\tan {}^{2}u}+{\cot }^{-1}\frac{1-\tan {}^{2}u}{2\tan u}=\frac{\pi }{3}$$$${\tan }^{-1}\tan 2u+{\cot }^{-1}\cot 2u=\frac{\pi }{3}$$$$2u+2u=\frac{\pi }{3}$$$$4u=\frac{\pi }{3}$$$$u=\frac{\pi }{12}$$$${\tan }^{-1}x=\frac{\pi }{12}$$$$x=\tan \frac{\pi }{12}$$$$=2-\sqrt{3}$$

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