Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Since $x\in (0,1)$, we have $\frac{x}{1-x}>0$, $\frac{1}{x^3}>0$, and $\frac{x}{1-x^2}\cdot \frac{1}{x^3}>1$.

Therefore:

$${\tan }^{-1}\left(\frac{x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{1}{x^3}\right)=\pi +{\tan }^{-1}\left(\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{1-x^2}\cdot \frac{1}{x^3}}\right)=\pi +{\tan }^{-1}\left(\frac{-1}{x}\right)=\frac{3\pi }{4}$$

This implies $x=1$, which is not possible given the condition $x\in (0,1)$.