Prove-that-arctan-x-2arctan-1-x-2-x-pi-2-

Question Number 94241 by Ar Brandon last updated on 17/May/20

Prove that

\arctan (x) + 2 \arctan (\sqrt{1 + x^{2}} - x) = \frac{π}{2}

Commented by mathmax by abdo last updated on 17/May/20

l e t φ (x) = a r c t a n x + 2 s r c t a n (\sqrt{1 + x^{2}} - x) \Rightarrow

φ^{^{'}} (x) = \frac{1}{1 + x^{2}} + 2 \times \frac{{(\sqrt{1 + x^{2}} - x)}^{^{'}}}{1 + {(\sqrt{1 + x^{2}} - x)}^{2}}

= \frac{1}{1 + x^{2}} + 2 \times \frac{\frac{x}{\sqrt{1 + x^{2}}} - 1}{1 + 1 + x^{2} - 2 x \sqrt{1 + x^{2}} + x^{2}}

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

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

= \frac{(1 + x^{2}) \sqrt{1 + x^{2}} - x (1 + x^{2}) + (1 + x^{2}) x - (1 + x^{2}) \sqrt{1 + x^{2}}}{(1 + x^{2}) \sqrt{1 + x^{2}} (1 + x^{2} - x \sqrt{1 + x^{2}})} = 0 \Rightarrow

φ (x) = c

c = φ (0) = 0 + 2 a r c t a n (1) = 2 \times \frac{π}{4} = \frac{π}{2} t h e e q u a l i t y i s p r o v e d .

Answered by som(math1967) last updated on 17/May/20

\tan^{- 1} x + {2 \tan}^{- 1} (\sqrt{1 + x^{2}} - x)

\tan^{- 1} x + \tan^{- 1} \frac{2 (\sqrt{1 + x^{2}} - x)}{1 - {(\sqrt{1 + x^{2}} - x)}^{2}}

\tan^{- 1} x + \tan^{- 1} \frac{2 (\sqrt{1 + x^{2}} - x)}{1 - 1 - x^{2} - x^{2} + 2 x \sqrt{1 + x^{2}}}

\tan^{- 1} x + \tan^{- 1} \frac{2 (\sqrt{1 + x^{2}} - x)}{2 x (\sqrt{1 + x^{2}} - x)}

\tan^{- 1} x + \tan^{- 1} \frac{1}{x}

\tan^{- 1} x + \cot^{- 1} x

= \frac{π}{2}