Question

The number of solutions of the equation $ta{n}^{-1}\left(\frac{x}{1-x^2}\right)+ta{n}^{-1}\left(\frac{1}{x^3}\right)=\frac{3\pi }{4}$ belonging to the interval (0,1) is

• A. \N
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A \N

$\because x\in (0,1)\Rightarrow \frac{x}{1-x}>0;\frac{1}{x^3}>0$ & $\frac{x}{1-x^2}.\frac{1}{x^3}>1$
$\therefore ta{n}^{-1}\frac{x}{1-x^2}+ta{n}^{-1}\frac{1}{x^3}=\pi +ta{n}^{-1}\left[\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{1-x^2}.\frac{1}{x^3}}\right]=\pi +ta{n}^{-1}\left(\frac{-1}{x}\right)=\frac{3\pi }{4}$
$\Rightarrow x=1$ (not possible)