Q186 / 270 · NDA 2022 (Phase 1) · Mathematics

If $\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{x}{3}\right)=\frac{\pi}{4}$, where $0 \lt x \lt 6$, then what is $x$ equal to?

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    5

Solution

$\begin{aligned} & \because \quad \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{x}{3}\right)=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1}\left(\frac{x}{3}\right)=\tan ^{-1} 1-\tan ^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow & \tan ^{-1} \frac{x}{3}=\tan ^{-1}\left(\frac{1-\frac{1}{2}}{1+\frac{1}{2}}\right) \\ \Rightarrow & \tan ^{-1} \frac{x}{3}=\tan ^{-1}\left(\frac{1}{3}\right) \\ \Rightarrow & \frac{x}{3}=\frac{1}{3} \Rightarrow x=1\end{aligned}$

Hence, the correct answer is option: 'A'.

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