Question

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

• A. $x=1\pm \sqrt{2}$
• B. $x=\sqrt{2}\pm \sqrt{3}$
• C. $x=3\pm 2\sqrt{2}$
• D. all value of $x$

Answer 1

The correct option is A $x=1\pm \sqrt{2}$

If we put $x=\tan \theta ,$ the given equality becomes ${\tan }^{-1}y=4\theta$

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

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

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

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