If $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$, then the value of $x$ is

If $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$, then the value of $x$ is
  1. $\frac{1}{3}$
  2. 2
  3. 3
  4. $\frac{1}{2}$

Solution

$\cot ^{-1} 7+\cot ^{-1} 8+\cot ^{-1} 18=\cot ^{-1} x$ $\Rightarrow \tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{18}\right)=\tan ^{-1}\left(\frac{1}{x}\right)$ $\ldots\left[\tan ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1} x\right]$ $\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\left(\frac{1}{7}\right)\left(\frac{1}{8}\right)}\right)+\tan ^{-1}\left(\frac{1}{18}\right)=\tan ^{-1}\left(\frac{1}{x}\right)$ $\begin{aligned} & \Rightarrow \tan ^{-1}\left(\frac{3}{11}\right)+\tan ^{-1}\left(\frac{1}{18}\right)=\tan ^{-1}\left(\frac{1}{x}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\left(\frac{3}{11}\right)\left(\frac{1}{18}\right)}\right)=\tan ^{-1}\left(\frac{1}{x}\right) \\ & \Rightarrow x=3\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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