Question

If ${\cot }^{-1}(x-1)-{\cot }^{-1}(x+1)=\frac{5\pi }{12}$, then the value of $\left(x^2+\frac{4}{x^2}\right)$, is

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (C)

We have ${\cot }^{-1}(x-1)-{\cot }^{-1}(x+1)=\frac{5\pi }{12}$
$\Rightarrow \left(\frac{\pi }{2}-{\tan }^{-1}(x-1)\right)-\left(\frac{\pi }{2}-{\tan }^{-1}(x+1)\right)=\frac{5\pi }{12}$
$\Rightarrow {\tan }^{-1}(x+1)-{\tan }^{-1}(x-1)=\frac{5\pi }{12}$
$\Rightarrow {\tan }^{-1}\frac{2}{x^2}=\frac{5\pi }{12}$
$\Rightarrow x^2=\frac{2}{2+\sqrt{3}}=2(2-\sqrt{3})$
Hence, $\left(x^2+\frac{4}{x^2}\right)=8$