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  • If the sum of all the solutions of <img alt="\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{3},-1<x<1, x \neq 0" src="https://entrancecorner.

If the sum of all the solutions of \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{3},-1<x<1, x \neq 0 is \alpha-\frac{4}{\sqrt{3}}, then \alpha is equal to

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

x \in(-1,1) \quad \tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)=2 \tan ^{-1} \mathrm{x}
x \in(0,1) \quad \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x
x \in(-1,0) \quad \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\mathrm{IT}+\tan ^{-1}\left(\frac{2 x}{1-\mathrm{x}^{2}}\right)=\mathrm{IT}+2 \tan ^{-1} \mathrm{x}

x \in(0,1) \quad 2 \tan ^{-1} x+2 \tan ^{-1} x=\pi / 3
                          \tan ^{-1} \mathrm{x}=\frac{\pi}{12}
                          \mathrm{x}=2-\sqrt{3}

x \in(-1,0) \quad 2 \tan ^{-1} x+I T+2 \tan ^{-1} x=\pi / 3
                            4 \tan ^{-1} x=\frac{-2 \pi}{3}
                            \tan ^{-1} \mathrm{x}=-\pi / 6
                          \mathrm{x}=-1 / \sqrt{3}
(2-\sqrt{3})+\left(-\frac{1}{\sqrt{3}}\right)=\alpha-\frac{4}{\sqrt{3}}
2-4 / \sqrt{3}=\alpha-4 / \sqrt{3}
\alpha=2
 

Posted by

Irshad Anwar

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