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Prove : $\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x.\cot x$=1

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  • $\cot (x+y)=\large\frac{\cot x+\cot y-1}{\cot y+\cot x}$
$\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x.\cot x$
$\cot x\cot 2x-[\cot 3x(\cot 2x+\cot x)]$
$\cot x.\cot 2x-[\cot(x+2x)(\cot 2x+\cot x)]$
$\cot x.\cot 2x-[(\cot x.\cot 2x-1)\large\frac{\cot 2x+\cot x}{\cot 2x+\cot x}]$
$\cot x-\cot 2x-\cot x\cot 2x+1$
Hence proved
answered Apr 17, 2014 by sreemathi.v

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