Let x = cos 2θ ⇒ θ = $\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}$ x ... 1
∴ $\sqrt{1+\mathit{x}}$ = $\sqrt{1+\mathit{c}\mathit{o}\mathit{s}2\mathit{\theta }}$ = $\sqrt{1+2\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }-1}$ = $\sqrt{2}$ cos θ
$\sqrt{1-\mathit{x}}$ = $\sqrt{1-\mathit{c}\mathit{o}\mathit{s}2\mathit{\theta }}$ = $\sqrt{1-1-2\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }}$ = $\sqrt{2}$ sin θ
Let y = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}|\frac{\sqrt{1+\mathit{x}}-\sqrt{1-\mathit{x}}}{\sqrt{1+\mathit{x}}+\sqrt{1-\mathit{x}}}|$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}|\frac{\sqrt{2}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }-\sqrt{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\sqrt{2}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\sqrt{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}|$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}|\frac{1-\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}|$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left\{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}-\mathit{\theta }\right)\right\}$
= $\frac{\mathit{\pi }}{4}$ - θ = $\frac{\mathit{\pi }}{4}$ - $\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}$ x From 1
∴ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $-\frac{1}{2}\left(-\frac{1}{\sqrt{1-{\mathit{x}}^2}}\right)$ = $\frac{1}{2\sqrt{1-{\mathit{x}}^2}}$