A geometry problem by Ralph Macarasig

From $\sin^2 + \cos^2 = 1$ , we know that $\sec^2 = \tan^2 + 1$ and $\csc^2 = \cot^2 + 1$

Then,

$\sqrt{\sec^2 \theta + \csc^2 \theta - \dfrac12 \sec \theta \csc \theta}$

$= \sqrt{\tan^2 \theta+ 1 + \cot^2 \theta + 1 - \frac{1}{2}\sqrt{(\tan^2 \theta + 1)(\cot^2 \theta + 1)} }$

$= \sqrt{\tan^2 \theta + \cot^2 \theta + 2 - \frac{1}{2}\sqrt{\tan^2 \theta\cot^2 \theta + \tan^2 \theta + \cot^2 \theta + 1} }$

$= \sqrt{\tan^2 \theta + \cot^2 \theta + 2 - \frac{1}{2}\sqrt{\tan^2 \theta + \cot^2 \theta + 2} }$

From $\tan \theta + \cot \theta = 4$ , squaring both sides,

$\tan^2 \theta + 2\tan \theta \cot \theta + \cot^2 \theta = 16$

$\tan^2 \theta + \cot^2 \theta = 14$

Therefore,

$\sqrt{\tan^2 \theta + \cot^2 \theta + 2 - \frac{1}{2}\sqrt{\tan^2 \theta + \cot^2 \theta + 2} }$

$= \sqrt{14 + 2 - \frac{1}{2}\sqrt{14 + 2}}$

$=\sqrt{16 - \frac{1}{2}\sqrt{16}}$

$= \sqrt{14} \approx\boxed{3.742}$