$\cos ^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\beta$

Applying Componendo and Dividendo, we get

$\frac{\sqrt{1+x^{2}}}{\sqrt{1-x^{2}}} &=\frac{\cos \beta+1}{\cos \beta-1} \\ \frac{1+x^{2}}{1-x^{2}} &=\left(\frac{\cos \beta+1}{\cos \beta-1}\right)^{2} \\ \frac{1+x^{2}}{1-x^{2}} &=\left(-\frac{2 \cos ^{2}\left(\frac{\beta}{2}\right)}{2 \sin ^{2}\left(\frac{\beta}{2}\right)}\right)^{2} \\ \frac{1+x^{2}}{1-x^{2}} &=\cot ^{4}\left(\frac{\beta}{2}\right) \\ 1-x^{2} &=x^{2} \tan ^{4}\left(\frac{\beta}{2}\right)+\tan ^{4}\left(\frac{\beta}{2}\right)$

$x^{2}+x^{2} \tan ^{4}\left(\frac{\beta}{2}\right)=1-\tan ^{4}\left(\frac{\beta}{2}\right) \\ x^{2}=\frac{1-\tan ^{4}\left(\frac{\beta}{2}\right)}{1+\tan ^{4}\left(\frac{\beta}{2}\right)} \\ x^{2}=\frac{\left[\sin ^{2}\left(\frac{\beta}{2}\right)-\cos ^{2}\left(\frac{\beta}{2}\right)\right]\left[\sin ^{2}\left(\frac{\beta}{2}\right)+\cos ^{2}\left(\frac{\beta}{2}\right)\right]}{\left[\sin ^{2}\left(\frac{\beta}{2}\right)+\cos ^{2}\left(\frac{\beta}{2}\right)\right]^{2}-2 \sin ^{2}\left(\frac{\beta}{2}\right) \cos ^{2}\left(\frac{\beta}{2}\right)} \\ x^{2}=\frac{-2 \cos \beta}{1+\cos ^{2} \beta}$

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Inverse Trigo. Functions

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