Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We analyze the equation based on the value of $x$. For $x\ge 1$, ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\pi -2{\tan }^{-1}x$ and ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x$. Substituting these, we get $(\pi -2{\tan }^{-1}x)-2{\tan }^{-1}x=\frac{\pi }{2}$, which simplifies to $4{\tan }^{-1}x=\frac{\pi }{2}$, so $x=\tan (\frac{\pi }{8})=\sqrt{2}-1$. This contradicts $x\ge 1$, so no solution in this range. For $-1\le x\le 1$, both terms are $2{\tan }^{-1}x$, so $2{\tan }^{-1}x-2{\tan }^{-1}x=\frac{\pi }{2}\Rightarrow 0=\frac{\pi }{2}$, which is false. For $x<-1$, ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=-\pi -2{\tan }^{-1}x$ and ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=-2{\tan }^{-1}x$. Substituting gives $(-\pi -2{\tan }^{-1}x)-(-2{\tan }^{-1}x)=\frac{\pi }{2}$, which simplifies to $-\pi =\frac{\pi }{2}$, also false. Thus, there is no solution to the given equation.