Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The equation is ${\tan }^{-1}(\frac{x}{1-x^2})+{\tan }^{-1}(\frac{1}{x^3})=\frac{3\pi }{4}$ for $x\in (0,1)$.

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

Also, $\frac{x}{1-x^2}\cdot \frac{1}{x^3}>1$.

Therefore,

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

$$\pi +{\tan }^{-1}(-\frac{1}{x})=\frac{3\pi }{4}⟹{\tan }^{-1}(-\frac{1}{x})=-\frac{\pi }{4}⟹-\frac{1}{x}=-1⟹x=1$$

Since $x=1$ is not in the interval $(0,1)$, there are no solutions.

Final Answer: The final answer is $\boxed{0}$