<s>[INST] If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$. If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.
I'm getting $\sec2A=\cfrac{25}{7}$. Is that correct?
 [/INST] Since $\sin A = \frac{3}{5} $, then $\cos A = \frac{4}{5} $. Therefore,
$$ \sec (2A) = \frac{1}{\cos(2A)} = \frac{1}{\cos^2 A - \sin^2 A} = \frac{1}{\frac{16}{25} - \frac{9}{25}} = \frac{1}{\frac{7}{25}} = \frac{25}{7}$$
</s>