If sec 2A = 1 / sin 13A, determine the value of A in degrees.

$\sec 2A = \dfrac{1}{\sin 13A}$

$\sec 2A = \csc 13A$

$\sec 2A = \sec (90^\circ - 13A)$
 

Hence,
$2A = 90^\circ - 13A$

$15A = 90^\circ$

$A = 6^\circ$
 

You can also do trial and error using the choices
$\sec 2A = \dfrac{1}{\sin 13A}$

$\dfrac{1}{\cos 2A} = \dfrac{1}{\sin 13A}$
 

Try A = 3°:   $\dfrac{1}{\cos 6^\circ} \ne \dfrac{1}{\sin 39^\circ}$   ←   not ok

Try A = 5°:   $\dfrac{1}{\cos 10^\circ} \ne \dfrac{1}{\sin 65^\circ}$   ←   not ok

Try A = 6°:   $\dfrac{1}{\cos 12^\circ} = \dfrac{1}{\sin 78^\circ}$   ←   ok!

Try A = 7°:   $\dfrac{1}{\cos 14^\circ} \ne \dfrac{1}{\sin 91^\circ}$   ←   not ok
 

Answer: A = 6°