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tan^-1{x/(1+√1-x²)} with respect to sin(2cot^-1√1+x∕√1-x)

$Let u = \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}}) = \tan^{- 1} (\frac{\sin θ}{1 + \sqrt{1 - \sin^{2} θ}}) [Put x = \sin θ \Rightarrow {θ = sin}^{- 1} x] = \tan^{- 1} (\frac{\sin θ}{1 + \cos θ}) = \tan^{- 1} (\frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}) = \tan^{- 1} (\tan \frac{θ}{2}) = \frac{θ}{2} = \frac{\sin^{- 1} x}{2} \Rightarrow \frac{d u}{d x} = \frac{1}{2 \sqrt{1 - x^{2}}} ... (1) Let v = \sin (2 \cot^{- 1} \frac{\sqrt{1 + x}}{\sqrt{1 - x}}) = \sin (2 \cot^{- 1} \frac{\sqrt{1 + \cos 2 θ}}{\sqrt{1 - \cos 2 θ}}) [Put x = \cos 2 θ] = \sin (2 \cot^{- 1} \frac{\sqrt{2 \cos^{2} θ}}{\sqrt{2 \sin^{2} θ}}) = \sin (2 \cot^{- 1} (\cot θ)) = \sin 2 θ = 1 - \cos^{2} 2 θ = 1 - (\cos (\cos^{- 1} x))^{2} = 1 - x^{2} \Rightarrow \frac{d v}{d x} = - 2 x ... (2) ∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{1}{2 \sqrt{1 - x^{2}}} \div (- 2 x) [Using (1) and (2)] = \frac{1}{2 \sqrt{1 - x^{2}}} \times \frac{- 1}{2 x} = \frac{- 1}{4 x \sqrt{1 - x^{2}}}$

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