Q69 / 180 · COMEDK 2014 · Mathematics
If $\cot ^{-1} \frac{1}{5}+\cot ^{-1} \frac{1}{3}-\cot ^{-1} \frac{4}{7}=\cot ^{-1} x$, then $x$ is equal to
- Correct answerA0
- B3
- C$\frac{1}{\sqrt{3}}$
- D$\infty$
Solution
Given, equation is
$$
\begin{aligned}
&\cot ^{-1} \frac{1}{5}+\cot ^{-1} \frac{1}{3}-\cot ^{-1} \frac{4}{7}=\cot ^{-1} x \\
&\Rightarrow \tan ^{-1} 5+\left(\tan ^{-1} 3-\tan ^{-1} \frac{7}{4}\right)=\tan ^{-1} \frac{1}{x} \\
&\Rightarrow \tan ^{-1} 5+\tan ^{-1}\left(\frac{3-\frac{7}{4}}{1+3 \times \frac{7}{4}}\right)=\tan ^{-1} \frac{1}{x} \\
&\Rightarrow \tan ^{-1} 5+\tan ^{-1}\left(\frac{1}{5}\right)=\tan ^{-1} \frac{1}{x} \\
&\Rightarrow \tan ^{-1} 5+\cot ^{-1} 5=\tan ^{-1} \frac{1}{x} \\
&\Rightarrow \tan ^{-1} \frac{1}{x}=\frac{\pi}{2} \\
&\Rightarrow \frac{1}{x} \tan {\frac{\pi}{2}}=\tan ^{2} \Rightarrow x=0
\end{aligned}
$$
Hence, the correct answer is option: 'A'.