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QJEEM23Ma2a8a: JEE Main 2023 Mathematics - Class 12 JEEM23Ma2a8a Solution: If the sum of all the solutions of , is , then is equal to…

If the sum of all the solutions of {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) + {\cot ^{ - 1}}\left( {{{1 - {x^2}} \over {2x}}} \right) = {\pi \over 3}, - 1 < x < 1,x \ne 0 , is α−43\alpha - {4 \over {\sqrt 3 }} , then α\alpha is equal to _____________.

Step-by-Step Verified Solution

Case-I

-1 < x < 0

tan⁡−1(2x1−x2)+π+tan⁡−1(2x1−x2)=π3\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}

tan⁡−12x1−x2=−π3\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{-\pi}{3}

2tan⁡−1x=−π32 \tan ^{-1} x=\frac{-\pi}{3}

tan⁡−1x=−π6\tan ^{-1} x=\frac{-\pi}{6}

x=−13x=\frac{-1}{\sqrt{3}}

Case-II

0 < x < 1

tan⁡−12x1−x2+tan⁡−12x1−x2=π3\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}

amp;tan⁡−12x1−x2=π6amp;2tan⁡−1x=π6amp;tan⁡−1x=π12amp;x=2−3amp; Sum =−13+2−3=2−43amp;⇒α=2\begin{aligned}& \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{6} \\\\& 2 \tan ^{-1} x=\frac{\pi}{6} \\\\& \tan ^{-1} x=\frac{\pi}{12} \\\\& x=2-\sqrt{3} \\\\& \text { Sum }=\frac{-1}{\sqrt{3}}+2-\sqrt{3}=2-\frac{4}{\sqrt{3}} \\\\& \Rightarrow \alpha=2\end{aligned}

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