Question

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

• A. $notdefined$
• B. $\sqrt{3}$
• C. $\pm 1$
• D. None of the above

Answer 1

Answer: (A)

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, $0<x<1$