$\frac{1}{\cos 69^\circ}$

Note that $\sin(69 ^\circ) = \cos(21 ^\circ)$

simplifying we get :

$(1- \csc ^2(21 ^\circ))(1-\sec^2(21 ^\circ))$

We now make use of the identities:

$\csc^2(x)-1= \cot ^2(x)$ and $\sec^2(x)-1=\tan^2(x)$ .

Simplifying,

$= (-1)( \csc ^2(21 ^\circ)-1)((\sec^2(21 ^\circ)-1)(-1)$

$= (\tan^2(21 ^\circ))(\cot^2(21 ^\circ))$

$\large = \boxed{\boxed{\boxed{1}}}.$

A clever pseudo-solution:

Note that there is nothing special about $21$ and $69$ except that they add up to $90$ . Therefore, we conjecture that any two $\alpha , \beta$ that satisfy $\alpha+\beta=90$ will give the same answer.

Now let $\alpha=\beta=45$ . Substituting, we get $(1-\sqrt{2})^2(1+\sqrt{2})^2=\boxed{1}$ and we are done.

i just figured out value would be positive ... then i made a guess :)