Question

Solve for $x,\,\,{{\tan }^{-1}}(1/x)=\pi +{{\tan }^{-1}}x,\,\,0 < x < 1$

• A. $not\, defined$
• B. $\sqrt{3}$
• C. $\pm \,1$
• D. None of the above

Answer 1

Answer: (A)

$not\, defined$

Answer 2

We have, ${{\tan }^{-1}}\left( \frac{1}{x} \right)=\pi +{{\tan }^{-1}}x,\,0 < x < 1$
$\Rightarrow$ ${{\tan }^{-1}}\left( \frac{1}{x} \right)-{{\tan }^{-1}}x=\pi$
$\Rightarrow$ ${{\tan }^{-1}}\left( \frac{\frac{1}{x}-x}{1+\frac{1}{x}.x} \right)=\pi$
$\Rightarrow$ ${{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{x(1+1)} \right)=\pi$
$\Rightarrow$ $\frac{1-{{x}^{2}}}{2x}=\tan \pi$
$\Rightarrow$ $\frac{1-{{x}^{2}}}{2x}=0$
$\Rightarrow$ $1-{{x}^{2}}=0\,\Rightarrow {{x}^{2}}=1\,\Rightarrow x=\pm 1$ but given, $0 < x < 1$