Solve the equation for exact solutions.
-$$\tan ^ { - 1 } ( x + 2 ) + \tan ^ { - 1 } x = \tan ^ { - 1 } \frac { 1 } { 4 }$$

Question

Quizplus · Accepted Answer

Using the formula for the sum of two inverse tangents, $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, we can set up the equation $\tan^{-1}\left(\frac{x+2+x}{1-(x+2)x}\right) = \tan^{-1}\left(\frac{1}{4}\right)$. Simplifying gives $\tan^{-1}\left(\frac{2x+2}{1-x^2-2x}\right) = \tan^{-1}\left(\frac{1}{4}\right)$, which implies $\frac{2x+2}{1-x^2-2x} = \frac{1}{4}$. Solving this equation for $x$ yields $x = -5 + 3\sqrt{2}$, which matches choice A.

Solve the Equation for Exact Solutions tan⁡−1(x+2)+tan⁡−1x=tan⁡−114\tan ^ { - 1 } ( x + 2 ) + \tan ^ { - 1 } x = \tan ^ { - 1 } \frac { 1 } { 4 }tan−1(x+2)+tan−1x=tan−141​

Solve the Equation for Exact Solutions $\tan ^ { - 1 } ( x + 2 ) + \tan ^ { - 1 } x = \tan ^ { - 1 } \frac { 1 } { 4 }$