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If Sin A = 3/5, and Cos A < 0, find Tan A
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\[\sin A=\frac{ 3 }{ 5 }>0,\] Hence A lies in 1st and 2nd quadrant. \[\cos A<0,\] hence A lies in 2nd and 3rd quadrant, combining we get A lies in 2nd quadrant \[\cos A=-\sqrt{1-\sin ^2 A}=-\sqrt{1-\left( \frac{ 3 }{ 5 } \right)^2}=-\frac{ 4 }{ 5}\] \[\tan A=\frac{ \sin A }{ \cos A }=\frac{ \frac{ 3 }{ 5 } }{ -\frac{ 4 }{ 5 } }=-\frac{ 3 }{ 4 }\]
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