# Question of inverse trigonometry.

How do I evaluate $\sin \left(\sec^{-1}\left(\frac{-7}{5}\right)\right)$?

The answer given is $\frac{2\sqrt{6}}{7}$, but I have no idea how to get it. Hope someone can explain it for me. Thanks in advance.

where the last line is due to Pythagoras Theorem. Also think of which quadrant is $\cos^{-1}\left( -\frac57 \right)$ located.