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

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(D) Given expression: $\frac{1}{	an 3A - 	an A} - \frac{1}{\cot 3A - \cot A}$Since $\cot 	heta = \frac{1}{	an 	heta}$,we have $\frac{1}{\cot 3A -

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$\cot 2A$

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(D) Given expression: $\frac{1}{	an 3A - 	an A} - \frac{1}{\cot 3A - \cot A}$Since $\cot 	heta = \frac{1}{	an 	heta}$,we have $\frac{1}{\cot 3A - \cot A} = \frac{1}{\frac{1}{	an 3A} - \frac{1}{	an A}} = \frac{	an A 	an 3A}{	an A - 	an 3A} = -\frac{	an A 	an 3A}{	an 3A - 	an A}$.Substituting this back into the expression:$= \frac{1}{	an 3A - 	an A} - (-\frac{	an A 	an 3A}{	an 3A - 	an A})$$= \frac{1 + 	an A 	an 3A}{	an 3A - 	an A}$Using the identity $	an(x - y) = \frac{	an x - 	an y}{1 + 	an x 	an y}$,we know that $	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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