SOLUTION : $\left.\begin{array}{l}0 \leq\left(\tan ^{-1} x\right)^2 \leq \frac{\pi^2}{4} \\ 0 \leq\left(\cos ^{-1} y\right)^2 \leq \pi^2\end{array}\right] \Rightarrow\left(\tan ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2 \leq \frac{5 \pi^2}{4}$
But $\left(\tan ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2=\pi^2 k$ hence $k \pi^2 \leq \frac{5 \pi^2}{4}, \quad k \leq \frac{5}{4}$
Now put $\tan ^{-1} x=\frac{\pi}{2}-\cos ^{-1} y\left(\frac{\pi}{2}-\cos ^{-1} y\right)^2+\left(\cos ^{-1} y\right)^2=\pi^2 k\left(\right.$ where $\left.\cos ^{-1} y=t\right)$
$2 \mathrm{t}^2-\pi \mathrm{t}+\left(\frac{\pi^2}{4}-\mathrm{k} \pi^2\right)=0$
For real roots, $D \geq 0$
$\pi^2-8\left(\frac{\pi^2}{4}-k \pi^2\right) \geq 0 \Rightarrow 1-2+8 k \geq 0, k \geq \frac{1}{8}$
From (i) and (ii), $k=1$
With $k=1, t=\frac{\pi \pm \sqrt{8 \pi^2-\pi^2}}{4}=\frac{\pi+\sqrt{7} \pi}{4}=(1 \pm \sqrt{7}) \frac{\pi}{4}$.
or $\quad \cos ^{-1} y=(\sqrt{7}+1) \frac{\pi}{4}\left(\right.$ as $\left.0 \leq \cos ^{-1} y \leq \pi\right) \quad \therefore \quad y=\cos (\sqrt{7}+1) \frac{\pi}{4}$
$\therefore \quad \tan ^{-1} x=\frac{\pi}{2}-(\sqrt{7}+1) \frac{\pi}{4}=\frac{\pi}{4}[(1-\sqrt{7})] \quad \Rightarrow \quad x=\tan (1-\sqrt{7}) \frac{\pi}{4} .$
