If θ is a positive acute such that sec θ = cosec 60°

We have: $\sec \theta=\operatorname{cosec} 60^{\circ}$ where $\theta$ is positive acute angle

$\Rightarrow \operatorname{cosec}\left(90^{\circ}-\theta\right)=\operatorname{cosec} 60^{\circ}$

$\Rightarrow 90^{\circ}-\theta=60^{\circ}$

$\Rightarrow \theta=30^{\circ}$

Now we have to find $2 \cos ^{2} \theta-1$

Put $\theta=30^{\circ}$

$=2 \times \cos ^{2} 30^{\circ}-1$

$=2 \times\left(\frac{\sqrt{3}}{2}\right)^{2}-1$

$=2 \times \frac{3}{4}-1$

$=\frac{1}{2}$

Hence the value of $2 \cos ^{2} \theta-1$ is $\frac{1}{2}$