## Calculus 8th Edition

$(secy-cosy)=tany$ $siny$
Need to prove the identity $secy-cosy=tany$ $siny$ In order to prove this, take left side isolate. $secy-cosy=\frac{1}{cosy}-cosy$ $=\frac{1-cos^{2}y}{cosy}$ Since, ${1-cos^{2}y}=sin^{2}y$, Thus, $secy-cosy=\frac{sin^{2}y}{cosy}$ Now, $secy-cosy=\frac{siny}{cosy}\times$ $siny$ Hence, $(secy-cosy)=tany$ $siny$