Question

Solution of ${\tan }^{-1}\frac{2x}{1-x^2}+{\cot }^{-1}\frac{1-x^2}{2x}=\frac{2\pi }{3}$ are

• A. $\frac{1}{\sqrt{3}}$
• B. $-\sqrt{3}$
• C. $\sqrt{3}+2$
• D. $\sqrt{3}-2$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

Given,

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

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

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

or, $4{\tan }^{-1}x=\frac{2\pi }{3}$

or, ${\tan }^{-1}x=\frac{\pi }{6}$

or, $x=\frac{1}{\sqrt{3}}$.