Verified Solution • 2025–26 Session moderate

QJEEM23Ma9773: JEE Main 2023 Mathematics - Class 12 JEEM23Ma9773 Solution: Let . If denotes the number of elements in then :

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 n(S)\mathrm{n(S)} denotes the number of elements in S\mathrm{S} then :

Multiple Choice Options

  • A n(S)=0\mathrm{n}(\mathrm{S})=0
  • B n(S)=1\mathrm{n}(\mathrm{S})=1 and only one element in S\mathrm{S} is less than 12\frac{1}{2} .
  • C n(S)=1\mathrm{n}(\mathrm{S})=1 and the elements in S\mathrm{S} is more than 12\frac{1}{2} .
  • D n(S)=1\mathrm{n}(\mathrm{S})=1 and the element in S\mathrm{S} is less than 12\frac{1}{2} .

Step-by-Step Verified Solution

 2tan⁡−1(1−x1+x)=cos⁡−1(1−x21+x2){\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}

amp; Put x=tan⁡θθ∈(0,π4)amp;2tan⁡−1(1−tan⁡θ1+tan⁡θ)=cos⁡−1(1−tan⁡2θ1+tan⁡2θ)amp;2tan⁡−1[tan⁡(π4−θ)]=cos⁡−1[cos⁡(2θ)]amp;⇒2(π4−θ)=2θ⇒θ=π8amp;⇒x=tan⁡π8=2−1≃0.414\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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