Question

If ${\tan }^{-1}y=4{\tan }^{-1}x(|x|<\tan \frac{\pi }{8})$, then which of the following options is (are) CORRECT?

• A. $\tan \frac{\pi }{8}$ is a root of $x^4+6x^2-1=0$
• B. $\tan \frac{\pi }{8}$ is a root of $x^4-6x^2+1=0$
• C. $\tan \frac{\pi }{16}$ is a root of $x^4+4x^3-6x^2-4x+1=0$
• D. $\tan \frac{\pi }{16}$ is a root of $x^4-4x^3-6x^2+4x+1=0$

Answer 1

Answer: (B), (C)

The correct options are
B $\tan \frac{\pi }{8}$ is a root of $x^4-6x^2+1=0$
C $\tan \frac{\pi }{16}$ is a root of $x^4+4x^3-6x^2-4x+1=0$

${\tan }^{-1}y=4{\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}y=2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ as $|x|<1$

Since, $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$
$\Rightarrow \tan \frac{\pi }{4}=\frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}=1$
$\Rightarrow 2\tan \frac{\pi }{8}=1-{\tan }^2\frac{\pi }{8}$
$\Rightarrow {(1+\tan \frac{\pi }{8})}^2=2$
Now, $|x|<\tan \frac{\pi }{8}$
$\Rightarrow (1+|x|{)}^2<2$
$\Rightarrow 1+x^2+2|x|<2$
$\Rightarrow 2|x|<1-x^2$
$\Rightarrow \left|\frac{2x}{1-x^2}\right|<1$
So, we can use the property
$2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2},\text{if}|x|<1$

$\therefore {\tan }^{-1}y={\tan }^{-1}\frac{\frac{4x}{1-x^2}}{1-\frac{4x^2}{(1-x^2{)}^2}}$
$\Rightarrow {\tan }^{-1}y={\tan }^{-1}\frac{4x(1-x^2)}{x^4-6x^2+1}$
$\Rightarrow y=\frac{4x(1-x^2)}{x^4-6x^2+1}$

If $x=\tan \frac{\pi }{8}$,
then ${\tan }^{-1}y=4{\tan }^{-1}x=\frac{\pi }{2}$
$\Rightarrow y=\frac{1}{0}=\frac{4x(1-x^2)}{x^4-6x^2+1}\Rightarrow x^4-6x^2+1=0$
Hence, $\tan \frac{\pi }{8}$ is a root of $x^4-6x^2+1=0$.

If $x=\tan \frac{\pi }{16}$,
then ${\tan }^{-1}y=4{\tan }^{-1}x=\frac{\pi }{4}$
$\Rightarrow y=1=\frac{4x(1-x^2)}{x^4-6x^2+1}\Rightarrow x^4+4x^3-6x^2-4x+1=0$
Hence, $\tan \frac{\pi }{16}$ is a root of $x^4+4x^3-6x^2-4x+1=0$.