If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.

Question

If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.

I'm getting $\sec2A=\cfrac{25}{7}$. Is that correct?

Answer 1

\begin{align} \sec2A&=\frac{1}{\cos2A}\\ &=\frac{1}{1-2\sin^2A}\\ &=\frac{1}{1-2\left(\frac{3}{5}\right)^2}\\ &=\frac{1}{1-2\left(\frac{9}{25}\right)}\\ &=\frac{1}{1-\frac{18}{25}}\\ &=\frac{1}{\frac{7}{25}}\\ &=\frac{25}{7} \end{align}

Answer 2

Since $\sin A = \frac{3}{5} $, then $\cos A = \frac{4}{5} $. Therefore,

$$ \sec (2A) = \frac{1}{\cos(2A)} = \frac{1}{\cos^2 A - \sin^2 A} = \frac{1}{\frac{16}{25} - \frac{9}{25}} = \frac{1}{\frac{7}{25}} = \frac{25}{7}$$