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Miscellaneous Exercise 3 · Q119

Q.Prove that tan⁡−1x=12cos⁡−1(1−x21+x2)\tan^{-1}x = \dfrac{1}{2}\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right), if x∈[0,1]x \in [0,1].

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Let x=tan⁡ϕx=\tan\phi with ϕ=tan⁡−1x∈[0,π4]\phi=\tan^{-1}x\in\left[0,\dfrac{\pi}{4}\right] (since x∈[0,1]x\in[0,1]). Then 1−x21+x2=1−tan⁡2ϕ1+tan⁡2ϕ=cos⁡2ϕ\dfrac{1-x^2}{1+x^2}=\dfrac{1-\tan^2\phi}{1+\tan^2\phi}=\cos2\phi (a standard identity). So cos⁡−1(1−x21+x2)=cos⁡−1(cos⁡2ϕ)=2ϕ\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)=\cos^{-1}(\cos2\phi)=2\phi, since 2ϕ∈[0,π2]⊂[0,π]2\phi\in\left[0,\dfrac{\pi}{2}\right]\subset[0,\pi], the principal range of cos⁡−1\cos^{-1}. So $\dfrac …

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