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If sin A =9/41 then find 1) tan A-cosec Aâ€‹

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$\sin A = \frac{9}{41}\\ \cos A = \sqrt{1 - \sin^2 A}\\\\ \cos A = \sqrt{1 -(\frac{9}{41})^2}\\\\ \cos A = \frac{40}{41}\\\\ \tan A - \csc A = \frac{\sin A}{\cos A} - \frac{1}{\sin A} \\\\ =\frac{9}{40} - \frac{41}{9} \\\\ =\frac{81-1640}{360}\\=\frac{-1559}{360}$

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Ravindra Pindel

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