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If $\theta=\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)$, then $\cot \theta$ is equal to
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Verified Answer
The correct answer is:
3
It is given that,
$\begin{aligned}
\theta & =\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18) \\
& =\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{18}\right) \\
& =\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7 \times 8}}\right)+\tan ^{-1} \frac{1}{18} \\
& =\tan ^{-1}\left(\frac{15}{55}\right)+\tan ^{-1}\left(\frac{1}{18}\right)=\tan ^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11 \times 18}}\right) \\
& =\tan ^{-1}\left(\frac{54+11}{195}\right)=\tan ^{-1}\left(\frac{1}{3}\right) \\
\Rightarrow & \tan \theta=\frac{1}{3} \Rightarrow \cot \theta=3
\end{aligned}$
Hence, option (2) is correct.
$\begin{aligned}
\theta & =\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18) \\
& =\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{18}\right) \\
& =\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7 \times 8}}\right)+\tan ^{-1} \frac{1}{18} \\
& =\tan ^{-1}\left(\frac{15}{55}\right)+\tan ^{-1}\left(\frac{1}{18}\right)=\tan ^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11 \times 18}}\right) \\
& =\tan ^{-1}\left(\frac{54+11}{195}\right)=\tan ^{-1}\left(\frac{1}{3}\right) \\
\Rightarrow & \tan \theta=\frac{1}{3} \Rightarrow \cot \theta=3
\end{aligned}$
Hence, option (2) is correct.
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