False

$\cos ^{2} 23^{\circ}-\sin ^{2} 67^{\circ}=\left(\cos 23^{\circ}-\sin 67^{\circ}\right)\left(\cos 23^{\circ}+\sin 67^{\circ}\right)\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]$

$=\left[\cos 23^{\circ}-\sin \left(90^{\circ}-23^{\circ}\right)\right]\left(\cos 23^{\circ}+\sin 67^{\circ}\right)$

$=\left(\cos 23^{\circ}-\cos 23^{\circ}\right)\left(\cos 23^{\circ}+\sin 67^{\circ}\right)\left[\because \sin \left(90^{\circ}-0\right)=\cos 0\right]$

$=0 \cdot\left(\cos 23^{\circ}+\sin 67^{\circ}\right)=0$

which may be either positive or negative.