If A, B, C are the angles of a triangle,then | vedclass

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(A) Given the expression $\sum \frac{\cot A + \cot B}{	an A + 	an B}$.We know that $\cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}

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(A) Given the expression $\sum \frac{\cot A + \cot B}{	an A + 	an B}$.We know that $\cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin B \cos A + \cos B \sin A}{\sin A \sin B} = \frac{\sin(A+B)}{\sin A \sin B}$.Also,$	an A + 	an B = \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \frac{\sin(A+B)}{\cos A \cos B}$.Therefore,$\frac{\cot A + \cot B}{	an A + 	an B} = \frac{\sin(A+B)}{\sin A \sin B} 	imes \frac{\cos A \cos B}{\sin(A+B)} = \cot A \cot B$.Summing over the cyclic permutations,we get $\sum \cot A \cot B = \cot A \cot B + \cot B \cot C + \cot C \cot A$.For any triangle $ABC$,$A+B+C = \pi$,which implies $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.

If $A, B, C$ are the angles of a triangle,then $\sum \frac{\cot A + \cot B}{\tan A + \tan B} = $

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