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