Question

Let $S =\left\{x \in R : 0< x< 1\right.$ and $\left.2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$. 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$