If ${\tan }^{-1}\mathrm{y}=4{\tan }^{-1}\mathrm{x}$, then 1/y is zero for

If we put x = tanθ, the given equality becomes tan^–1y = 4θ

$\Rightarrow \mathrm{y}=\tan 4\mathrm{\theta }=\frac{2\tan 2\mathrm{\theta }}{1-{\tan }^{2}2\mathrm{\theta }}=\frac{2\left[\frac{2\tan \mathrm{\theta }}{1-{\tan }^2\mathrm{\theta }}\right]}{1-{\left(\frac{2\tan \mathrm{\theta }}{1-{\tan }^2\mathrm{\theta }}\right)}^2}$

$=\frac{2\times 2\mathrm{x}\left(1-{\mathrm{x}}^2\right)}{{\left(1-{\mathrm{x}}^2\right)}^2-4{\mathrm{x}}^2}=\frac{4\mathrm{x}\left(1-{\mathrm{x}}^2\right)}{1-6{\mathrm{x}}^2+{\mathrm{x}}^4}$

$\therefore \frac{1}{\mathrm{y}}\text{ is zero if }{\mathrm{x}}^4-6{\mathrm{x}}^2+1=0$

$\Rightarrow {\mathrm{x}}^2=\frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2}=(1\pm \sqrt{2}{)}^2$