Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The domain of ${\tan }^{-1}\left(\frac{1}{2x+1}\right)$ requires $2x+1\ne 0$, so $x\ne -\frac{1}{2}$. Case 1: If ${\left(\frac{x}{x+1}\right)}^2<1⟹x^2<(x+1{)}^2⟹0<2x+1⟹x>-\frac{1}{2}$. In this case, $2{\tan }^{-1}\left(\frac{x}{x+1}\right)={\tan }^{-1}\left(\frac{2\frac{x}{x+1}}{1-(\frac{x}{x+1}{)}^2}\right)={\tan }^{-1}\left(\frac{2x(x+1)}{(x+1{)}^2-x^2}\right)={\tan }^{-1}\left(\frac{2x(x+1)}{2x+1}\right)$. The equation becomes ${\tan }^{-1}\left(\frac{2x(x+1)}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{2x+1}\right)=\frac{\pi }{2}$.

For the sum of two ${\tan }^{-1}$ terms to be $\frac{\pi }{2}$, their arguments must be positive and their product must be 1. Since $x>-\frac{1}{2}$, $2x+1>0$, so $\frac{1}{2x+1}>0$. For $\frac{2x(x+1)}{2x+1}>0$, we need $x(x+1)>0$, which implies $x\in (0,\infty )$ (considering $x>-\frac{1}{2}$). Therefore, for $x\in (0,\infty )$, both arguments are positive. Setting their product to 1: $\left(\frac{2x(x+1)}{2x+1}\right)\left(\frac{1}{2x+1}\right)=1⟹2x(x+1)=(2x+1{)}^2$.$2x^2+2x=4x^2+4x+1⟹2x^2+2x+1=0$. The discriminant is $D=2^2-4(2)(1)=4-8=-4<0$. Thus, there are no real solutions in this case. Case 2: If ${\left(\frac{x}{x+1}\right)}^2>1⟹x<-\frac{1}{2}$. Specifically, if $x<-1$, then $\frac{x}{x+1}>1$. So $2{\tan }^{-1}\left(\frac{x}{x+1}\right)=\pi +{\tan }^{-1}\left(\frac{2x(x+1)}{2x+1}\right)$. The equation becomes $\pi +{\tan }^{-1}\left(\frac{2x(x+1)}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{2x+1}\right)=\frac{\pi }{2}$.${\tan }^{-1}\left(\frac{2x(x+1)}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{2x+1}\right)=-\frac{\pi }{2}$.

For $x<-1$, both $\frac{2x(x+1)}{2x+1}$ and $\frac{1}{2x+1}$ are negative. For their sum to be $-\frac{\pi }{2}$, their product must be 1. This again leads to $2x^2+2x+1=0$, which has no real solutions. Case 3: If $x=-\frac{1}{2}$, the term ${\tan }^{-1}\left(\frac{1}{2x+1}\right)$ is undefined. So $x=-\frac{1}{2}$ is not a solution. Combining all cases, there are no real solutions to the given equation. The final answer is $\boxed{0}$.