JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 14)
Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$.
If $$\mathrm{n(S)}$$ denotes the number of elements in $$\mathrm{S}$$ then :
$$\mathrm{n}(\mathrm{S})=0$$
$$\mathrm{n}(\mathrm{S})=1$$ and only one element in $$\mathrm{S}$$ is less than $$\frac{1}{2}$$.
$$\mathrm{n}(\mathrm{S})=1$$ and the elements in $$\mathrm{S}$$ is more than $$\frac{1}{2}$$.
$$\mathrm{n}(\mathrm{S})=1$$ and the element in $$\mathrm{S}$$ is less than $$\frac{1}{2}$$.
Explanation
$$ {\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}$$
$\begin{aligned} & \text { Put } x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\\\ & 2 \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\cos ^{-1}[\cos (2 \theta)] \\\\ & \Rightarrow 2\left(\frac{\pi}{4}-\theta\right)=2 \theta \Rightarrow \theta=\frac{\pi}{8} \\\\ & \Rightarrow x=\tan \frac{\pi}{8}=\sqrt{2}-1 \simeq 0.414\end{aligned}$
$\begin{aligned} & \text { Put } x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\\\ & 2 \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\cos ^{-1}[\cos (2 \theta)] \\\\ & \Rightarrow 2\left(\frac{\pi}{4}-\theta\right)=2 \theta \Rightarrow \theta=\frac{\pi}{8} \\\\ & \Rightarrow x=\tan \frac{\pi}{8}=\sqrt{2}-1 \simeq 0.414\end{aligned}$
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