Answer
$(secy-cosy)=tany$ $siny$
Work Step by Step
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, $(secy-cosy)=tany$ $siny$
