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The value of $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)$ is
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Verified Answer
The correct answer is:
$\frac{6}{17}$
$\frac{6}{17}$
$$
\begin{aligned}
& \text { Let } E=\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1} \frac{17}{6}\right)=\frac{6}{17} .
\end{aligned}
$$
\begin{aligned}
& \text { Let } E=\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1} \frac{17}{6}\right)=\frac{6}{17} .
\end{aligned}
$$
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