Question

If ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi }{3};xϵ(0,1).$ then $(x^4+\frac{1}{x^4})$ is equal to

• A. $194$
• B. $196$
• C. $192$
• D. $200$

Answer 1

The correct option is A $194$

We have,

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

$⟹{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

$⟹\left(\frac{2x}{1-x^2}\right)=\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$

$\frac{1-x^2}{2x}=\sqrt{3}⟹\frac{1-x^2}{x}=2\sqrt{3}$

$⟹\frac{1}{x}-x=2\sqrt{3}$ (Squaring)

$⟹x^2+\frac{1}{x^2}-2=12⟹x^2+\frac{1}{x^2}=14$ (Squaring again)

$⟹x^4+\frac{1}{x^4}+2=196$

$x^4+\frac{1}{x^4}=194$

Hence, this is the answer.