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Question Number 58500 by Kunal12588 last updated on 24/Apr/19

$$changeinsimplestform:$$$${\tan }^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$$

Answered by MJS last updated on 24/Apr/19

$$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{{(\sqrt{a}+\sqrt{b})}^2}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{a+2\sqrt{ab}+b}{a-b}$$$$\frac{1+x^2+2\sqrt{(1+x^2)(1-x^2)}+1-x^2}{1+x^2-(1-x^2)}=$$$$=\frac{1+\sqrt{1-x^4}}{x^2}$$$$\arctan \frac{1+\sqrt{1-x^4}}{x^2}=t$$$$\Rightarrow x=\pm \sqrt{\sin 2t}\Rightarrow t=\frac{2n\pi +\arcsin x^2}{2}\vee t=\frac{(2n+1)\pi -\arcsin x^2}{2}$$$$\mathrm{testing}\mathrm{values}\mathrm{we}\mathrm{get}$$$$\arctan \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi -\arcsin x^2}{2}$$

Commented by Kunal12588 last updated on 24/Apr/19

$$thankyousir,Ihavealsowrittenanans$$$$plscheck.$$

Answered by Kunal12588 last updated on 24/Apr/19

$${tan}^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$$$$let{x}^2=cosy$$$$\Rightarrow {cos}^{-1}{x}^2=y$$$${tan}^{-1}\frac{\sqrt{1+cosy}+\sqrt{1-cosy}}{\sqrt{1+cosy}-\sqrt{1-cosy}}$$$$={tan}^{-1}\frac{\sqrt{2}cos\frac{y}{2}+\sqrt{2}sin\frac{y}{2}}{\sqrt{2}cos\frac{y}{2}-\sqrt{2}sin\frac{y}{2}}$$$$={tan}^{-1}\frac{1+tan\frac{y}{2}}{1-tan\frac{y}{2}}$$$$={tan}^{-1}\frac{tan\frac{\pi }{4}+tan\frac{y}{2}}{1-tan\frac{\pi }{4}tan\frac{y}{2}}$$$$={tan}^{-1}\left(tan(\frac{\pi }{4}+\frac{y}{2})\right)$$$$=\frac{\pi }{4}+\frac{y}{2}$$$$=\frac{\pi }{4}+\frac{1}{2}{cos}^{-1}{x}^2$$

Commented by MJS last updated on 24/Apr/19

$$\mathrm{good}!$$$$\frac{\pi }{4}+\frac{1}{2}\arccos x^2=\frac{\pi }{2}-\frac{1}{2}\arcsin x^2$$

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