Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Using the identity ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$ (valid for $AB<1$), the equation becomes ${\tan }^{-1}\left(\frac{(x-1)+(x+1)}{1-(x-1)(x+1)}\right)={\tan }^{-1}(2)$. This simplifies to $\frac{2x}{1-(x^2-1)}=2$, so $\frac{2x}{2-x^2}=2$. Assuming $2-x^2\ne 0$, we get $x=2-x^2$, which leads to $x^2+x-2=0$. Factoring yields $(x+2)(x-1)=0$, so $x=1$ or $x=-2$. We must check the condition $(x-1)(x+1)<1$. For $x=1$, $(1-1)(1+1)=0<1$, so $x=1$ is a valid solution. For $x=-2$, $(-2-1)(-2+1)=(-3)(-1)=3$, which violates $AB<1$, so $x=-2$ is not a valid solution. Therefore, the only integer solution is $x=1$.