Let S=xR: 0

Question

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

Accepted Answer

Let $\tan^{-1}x=\theta\in\left(0,\dfrac{\pi}{4}\right)$, so $x=\tan\theta$ and $\dfrac{1-x}{1+x}=\tan\left(\dfrac{\pi}{4}-\theta\right)$.

$$2\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right)=\cos^{-1}(\cos2\theta)\ \Rightarrow\ 2\left(\frac{\pi}{4}-\theta\right)=2\theta\ \Rightarrow\ 4\theta=\frac{\pi}{2}\ \Rightarrow\ \theta=\frac{\pi}{8}.$$

So $x=\tan\dfrac{\pi}{8}=\sqrt2-1\approx0.414$, a single value less than $\dfrac12$. Hence $n(S)=1$, and the element is less than $\dfrac12$.

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