The real values of $x$ that satisfy the equation $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$ is
The real values of $x$ that satisfy the equation $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$ is
- $\frac{-3 \pm \sqrt{17}}{4}$
- $-1 \pm \sqrt{3}$
- $\sqrt{3}-1$
- $\frac{\sqrt{17}-3}{4}$
Solution
$\begin{aligned} & \text { } \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} \Rightarrow \tan ^{-1} \frac{x+2 x}{1-2 x^2}=\frac{\pi}{4} \\ & \Rightarrow \frac{x+2 x}{1-2 x^2}=1 \Rightarrow 2 x^2+3 x-1=0 \\ & \Rightarrow x=\frac{-3 \pm \sqrt{17}}{4} \Rightarrow x=\frac{-3+\sqrt{17}}{4} \\ & {\left[\because x=\frac{-3-\sqrt{17}}{4} \text { does not satisfy the equation }\right]}\end{aligned}$
Asked in: AP EAMCET 2024 (21 May Shift 2)
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