Question

Let $S=\{x\in R:0<x<1$ and $2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$. If $n(S)$ denotes the number of elements in $S$ then :

• A. $n(S)=2$ and only one element in $S$ is less than $\frac{1}{2}$.
• B. $n(S)=1$ and the element in $S$ is less than $\frac{1}{2}$.
• C. $n(S)=1$ and the elements in $S$ is more than $\frac{1}{2}$.
• D. $n(S)=0$

Answer 1

Answer: (B)

$0<x<1$
$2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
${\tan }^{-1}x=\theta \in \left(0,\frac{\pi }{4}\right)$
$\therefore x=\tan \theta$
$2{\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\theta \right)\right)={\cos }^{-1}(\cos 2\theta )$
$2\left(\frac{\pi }{4}-\theta \right)=2\theta$
$\therefore 4\theta =\frac{\pi }{2}\therefore \theta =\frac{\pi }{8}$
$x=\tan \frac{\pi }{8}$
$\therefore x=\sqrt{2}-1\simeq 0.414$