If csc A + cot A = frac{11}{2}, then tan A = | vedclass

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(C) We are given that $\csc A + \cot A = \frac{11}{2}$.We know the identity $\csc^2 A - \cot^2 A = 1$,which can be factored as $(\csc A - \cot A)(\csc

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$\frac{44}{117}$

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(C) We are given that $\csc A + \cot A = \frac{11}{2}$.We know the identity $\csc^2 A - \cot^2 A = 1$,which can be factored as $(\csc A - \cot A)(\csc A + \cot A) = 1$.Substituting the given value: $(\csc A - \cot A) 	imes \frac{11}{2} = 1$.Thus,$\csc A - \cot A = \frac{2}{11}$.Now,subtract the two equations:$(\csc A + \cot A) - (\csc A - \cot A) = \frac{11}{2} - \frac{2}{11}$.$2 \cot A = \frac{121 - 4}{22} = \frac{117}{22}$.$\cot A = \frac{117}{44}$.Since $	an A = \frac{1}{\cot A}$,we have $	an A = \frac{44}{117}$.

If $\csc A + \cot A = \frac{11}{2},$ then $\tan A = $

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