$\tan \theta = 4/8t$

$\tan \theta = \dfrac{1}{2t}$

$\sec^2 \theta \, \dfrac{d\theta}{dt} = \dfrac{-1(2)}{(2t)^2}$

$\dfrac{1}{\cos^2 \theta} \, \dfrac{d\theta}{dt} = -\dfrac{1}{2t^2}$

$\dfrac{d\theta}{dt} = -\cos^2 \theta \, \dfrac{1}{2t^2}$

after t = 1 sec

$8t = 8(1) = 8$

$\text{hypotenuse} = \sqrt{8^3 + 4^2 \,} = 4\sqrt{5}$

$\cos \theta = \dfrac{8}{4\sqrt{5}}$

$\dfrac{d\theta}{dt} = -\left( \dfrac{8}{4\sqrt{5}} \right)^2 \, \dfrac{1}{2(1^2)}$

$\dfrac{d\theta}{dt} = -\left( \dfrac{64}{80} \right) \, \dfrac{1}{2}$

$\dfrac{d\theta}{dt} = -\left( \dfrac{64}{80} \right) \, \dfrac{1}{2}$

$\dfrac{d\theta}{dt} = -\dfrac{2}{5} \, \text{ rad/sec}$ (*okay!*)