If {tan ^{ - 1}}x + 2{cot ^{ - 1}}x = frac{{2 | vedclass

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(C) Given equation: ${	an ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ We know that ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$ for all

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$\sqrt 3 $

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(C) Given equation: ${	an ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ We know that ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$ for all $x \in \mathbb{R}$. We can rewrite the given equation as: $({	an ^{ - 1}}x + {\cot ^{ - 1}}x) + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ Substituting the identity: $\frac{\pi }{2} + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ Subtracting $\frac{\pi }{2}$ from both sides: ${\cot ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{2}$ ${\cot ^{ - 1}}x = \frac{{4\pi - 3\pi }}{6} = \frac{\pi }{6}$ Taking the cotangent of both sides: $x = \cot \left( \frac{\pi }{6} ight)$ Since $\cot \left( \frac{\pi }{6} ight) = \sqrt 3 $,we get $x = \sqrt 3 $.

If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$

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