frac{1}{{tan 3A - tan A}} - frac{1}{{cot 3A - | vedclass

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(D) Given expression: $\frac{1}{{	an 3A - 	an A}} - \frac{1}{{\cot 3A - \cot A}}$We know that $\cot 	heta = \frac{1}{	an 	heta}$. Substituting th

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$\cot 2A$

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(D) Given expression: $\frac{1}{{	an 3A - 	an A}} - \frac{1}{{\cot 3A - \cot A}}$We know that $\cot 	heta = \frac{1}{	an 	heta}$. Substituting this into the second term:$\frac{1}{{\cot 3A - \cot A}} = \frac{1}{{\frac{1}{	an 3A} - \frac{1}{	an A}}} = \frac{1}{{\frac{	an A - 	an 3A}{	an 3A 	an A}}} = \frac{	an 3A 	an A}{	an A - 	an 3A} = -\frac{	an 3A 	an A}{	an 3A - 	an A}$Substituting this back into the original expression:$= \frac{1}{{	an 3A - 	an A}} - \left( -\frac{	an 3A 	an A}{	an 3A - 	an A} ight)$$= \frac{1 + 	an 3A 	an A}{	an 3A - 	an A}$Using the identity $	an(X - Y) = \frac{	an X - 	an Y}{1 + 	an X 	an Y}$,we have $	an(3A - A) = \frac{	an 3A - 	an A}{1 + 	an 3A 	an A}$.Therefore,$\frac{1 + 	an 3A 	an A}{	an 3A - 	an A} = \frac{1}{	an(3A - A)} = \frac{1}{	an 2A} = \cot 2A$.

$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $

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