Answer
$-2$
Work Step by Step
Find $\frac{dy}{d\theta}$:
$\frac{dy}{d\theta}=\frac{d}{d\theta}(r\sin\theta)=\frac{d}{d\theta}((\sin\theta+2\cos\theta)\sin\theta)$
$=\frac{d}{d\theta}(\sin\theta+2\cos\theta)\cdot \sin\theta+(\sin \theta+2\cos\theta)\frac{d}{d\theta}(\sin\theta)$
$=(\cos\theta-2\sin\theta)\sin\theta+(\sin\theta+2\cos\theta)\cos\theta$
$=\cos\theta\sin\theta-2\sin^2\theta+\cos\theta\sin\theta+2\cos^2\theta$
$=2\cos\theta\sin\theta-2(\cos^2\theta-\sin^2\theta)$
$=\sin2\theta-2\cos2\theta$
Find $\frac{dx}{d\theta}$:
$\frac{dx}{d\theta}=\frac{d}{d\theta}(r\cos\theta)=\frac{d}{d\theta}((\sin\theta+2\cos\theta)\cos\theta)$
$=\frac{d}{d\theta}(\sin\theta+2\cos\theta)\cos\theta+(\sin\theta+2\cos\theta)\frac{d}{d\theta}(\cos\theta)$
$=(\cos\theta-2\sin\theta)\cos\theta+(\sin\theta+2\cos\theta)(-\sin\theta)$
$=(\cos^2\theta-\sin^2\theta)-2(2\sin\theta\cos\theta)$
$=\cos2\theta-2\sin2\theta$
Find the slope of the tangent line at $\theta=\pi/2$:
$m=\frac{dy}{dx}|_{\theta=\pi/2}=\frac{dy/d\theta}{dx/d\theta}|_{\theta=\pi/2}=\frac{\sin\pi-2\cos\pi}{\cos\pi-2\sin\pi}=\frac{0-2(-1)}{-1-2\cdot0}=\frac{2}{-1}=-2$
Thus, the slope is $-2$.
