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Question Number 74526 by Kunal12588 last updated on 25/Nov/19

$$provethat$$$$\frac{1}{2}{tan}^{-1}x={cos}^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right)$$$$usingsubstitutionx=cos2\theta$$

Answered by mind is power last updated on 25/Nov/19

$$missclick$$$$ithinkisx=tg\left(2\theta \right)$$$$trueforx⩾0$$$$x=-1{tan}^{-}(-1)=-\frac{\pi }{4}$$$$weget-\frac{\pi }{8}$$$${cos}^{-}\left(x\right)\in [0,\pi ]$$$$sincet\to tg\left(2\theta \right)isbjection$$$$I=]0,\frac{\pi }{4}[\to {\mathbb{R}}^{+}$$$$\Rightarrow \mathrm{\forall }x\in \mathbb{R}\mathrm{\exists }!\theta \in I\mid x=tg\left(2\theta \right)$$$${\tan }^{-1}\left(tg\left(2\theta \right)\right)=2\theta since2\theta \in ]0,\frac{\pi }{2}[$$$$lHs=\theta$$$$1+x^2=\frac{1}{{cos}^2\left(2\theta \right)}\Rightarrow \sqrt{1+x^2}=\frac{1}{cos\left(2\theta \right)},sincecos\left(2\theta \right)⩾0overI$$$$\sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}}=\sqrt{\frac{1+\frac{1}{cos\left(2\theta \right)}}{\frac{2}{cos\left(2\theta \right)}}}=\sqrt{\frac{1+cos\left(2\theta \right)}{2}}$$$$1+cos\left(2\theta \right)=2{cos}^2\left(\theta \right)\Rightarrow \sqrt{\frac{2{cos}^2\left(\theta \right)}{2}}=\mid cos\left(\theta \right)\mid =cos\left(\theta \right)$$$${cos}^{-}\left(cos\left(\theta \right)\right)=\theta cause\theta \in ]0,\frac{\pi }{4}[$$$$so\theta =\theta this\Rightarrow \mathrm{\forall }x⩾0\frac{{tan}^{-}\left(x\right)}{2}={\cos }^{-1}\left(\frac{\sqrt{1+\sqrt{1+x^2}}}{2\sqrt{1+x^2}}\right)$$

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