Answer
(a)$\frac{d\theta}{dt}= \frac{ \cos^{2} \theta}{x^2}\left(x\frac{dy}{dt}-y \frac{dx}{dt}\right)$
(b)$-\frac{5}{16} \frac{rad}{s}$, decreasing
Work Step by Step
(a) $\tan\theta =\frac{y}{x}$
$ \frac{d(\tan \theta)}{dt} \frac{d\theta}{dt}=\frac{d(yx^{-1})}{dt}$
$ \sec^2 \theta \frac{d\theta}{dt}=x^{-1} \frac{dy}{dt}+y \frac{ dx^{-1}}{dt}$
$ \frac{1}{\cos^2 \theta}\frac{d\theta}{dt}= \frac{1}{x} \frac{dy}{dt}+y \frac{ dx^{-1}}{dx} \frac{dx}{dt}$
$ \frac{d\theta}{dt}= \cos^2\theta[\frac{1}{x} \frac{dy}{dt}+y \frac{ dx^{-1}}{dx} \frac{dx}{dt}]$
$ \frac{d\theta}{dt}= \cos^2\theta[\frac{1}{x} \frac{dy}{dt}+y(-1x^{-2} \frac{dx}{dt}]$
$ \frac{d\theta}{dt}= \cos^2\theta[\frac{1}{x} \frac{dy}{dt}- \frac{y}{x^2} \frac{dx}{dt}]$
$ \frac{d\theta}{dt}= \frac{ \cos^2\theta }{x^2}[\frac{1}{x} \frac{dy}{dt}- \frac{y}{x^2} \frac{dx}{dt}] x^2$
$ \frac{d\theta}{dt}= \frac{ \cos^2\theta }{x^2}[x \frac{dy}{dt}- y \frac{dx}{dt}] $
(b) $ x=2$ units ,$\frac{dx}{dt}=$ 1 unit/s $y=2$ units ,$ \frac{dy}{dt}=-\frac{1}{4}$ units/s
Also
$ \cos^2\theta=\frac{1}{\sec^\theta}= \frac{1}{1+\tan^2\theta}= \frac{1}{1+\frac{y^2}{x^2}} $
$ \cos^2\theta= \frac{x^2}{x^2+y^2} $
Put $x=2,y=2$
$ \cos^2\theta=\frac{2^2}{2^2+2^2}=\frac{4}{8}=\frac{1}{2}$
Putting $ x=2,y=2, \cos^2 \theta=\frac{1}{2}, \frac{dx}{dt}=1, \frac{dy}{dt}=-\frac{1}{4}$ in equation
$ \frac{d\theta}{dt}= \frac{ \cos^2\theta }{x^2}[x \frac{dy}{dt}- y \frac{dx}{dt}] $
$ \frac{d\theta}{dt}= \frac{1}{2} \frac{1}{4} [ -2(\frac{1}{4})-2 (1)=-\frac{5}{16} \frac{rad}{s}$
Therefore $\theta$ is decreasing.
