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The value of $\cot 2 x \cot x-\cot 3 x \cot 2 x$
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1
We have, $\cot 3 x=\cot (2 x+x)$
$\begin{aligned}
&\Rightarrow \cot 3 x=\frac{\cot 2 x \cot x-1}{\cot 2 x+\cot x} \\
&\left\{\cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}\right\} \\
&\Rightarrow \cot 3 x \cot 2 x+\cot 3 x \cot x=\cot 2 x \cot x-1 \\
&\Rightarrow \cot 2 x \cot x-\cot 3 x \cot 2 x-\cot 3 x \cot x=1
\end{aligned}$
##0
$\begin{aligned}
&\Rightarrow \cot 3 x=\frac{\cot 2 x \cot x-1}{\cot 2 x+\cot x} \\
&\left\{\cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}\right\} \\
&\Rightarrow \cot 3 x \cot 2 x+\cot 3 x \cot x=\cot 2 x \cot x-1 \\
&\Rightarrow \cot 2 x \cot x-\cot 3 x \cot 2 x-\cot 3 x \cot x=1
\end{aligned}$
##0
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