Hi Diana,

I don't memorize many trig formulas but I do know

\[ \sin^2 x + \cos^2 x = 1\]

and the double angle formulas for the sine and cosine, in particuar

\[\cos 2x = \cos^2 x - \sin^2 x.\]

Substituting from the first equation into the second I get

\[\cos 2x = 1 + 2 \cos^2 x\]

and hence

\[\cos x = \sqrt{ \left( \frac{cos 2x - 1}{2} \right)} .\]

Since

\[\sec x = \frac{1}{\cos x}\]

you can use the expression for the $\cos x$ above to determine the uantity you want if you know the cosine of $\large \frac{\pi}{6}.$

Harley