We know that:
$\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{2\mathit{x}}{1+{\mathit{x}}^2}$ = 2 $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x for |x| ≤ 1 .. (1)
$\mathit{c}\mathit{o}{\mathit{s}}^{-1}\frac{1-{\mathit{y}}^2}{1+{\mathit{y}}^2}$ = 2 $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ y for y = 0 ... (2)
∴ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{2\mathit{x}}{1+{\mathit{x}}^2}$ + $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\frac{1-{\mathit{y}}^2}{1+{\mathit{y}}^2}$ = $2\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x + 2 $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ y
⇒ tan $\frac{1}{2}|\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{2\mathit{x}}{1+{\mathit{x}}^2}+\mathit{c}\mathit{o}{\mathit{s}}^{-1}\frac{1-{\mathit{y}}^2}{1+{\mathit{y}}^2}|$
= tan $\frac{1}{2}$ (2 $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x + 2 $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ y)
= tan $\left(\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{y}\right)$
= tan $\left(\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\mathit{x}+\mathit{y}}{1-\mathit{x}\mathit{y}}\right)$
[Since $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{y}$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\mathit{x}+\mathit{y}}{1-\mathit{x}\mathit{y}}$ , for xy < 1]
= $\frac{\mathit{x}+\mathit{y}}{1-\mathit{x}\mathit{y}}$
OR
We know that:
$\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{y}$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\mathit{x}+\mathit{y}}{1-\mathit{x}\mathit{y}}$ , for xy < 1
We have:
$\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{2}\right)$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{5}\right)$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{8}\right)$
= $|\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{2}\right)$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{5}\right)|$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{8}\right)$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2}\times \frac{1}{5}}\right)$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{8}\right)$ (Since $\frac{1}{2}\times \frac{1}{5}$ <1)
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{7}{9}\right)+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{8}\right)$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{9}\times \frac{1}{8}}$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{56+9}{72-7}$ (Since $\frac{7}{9}\times \frac{1}{8}$ < 1)
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{65}{65}$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ 1 = $\frac{\mathit{\pi }}{4}$
Hence, $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{2}\right)$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{5}\right)$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{1}{8}\right)$ = $\frac{\mathit{\pi }}{4}$