Answer
(a) $\dfrac{\pi}{360}$
(b) $1+\dfrac{\pi}{90}$
Work Step by Step
(a) Since
\begin{align*}
\left.\frac{d}{d \theta} \sin \theta\right|_{\theta=60^{\circ}}&=\left.\frac{d}{d \theta} \sin \left(\frac{\pi}{180} \theta\right)\right|_{\theta=60^{\circ}}\\
&=\left(\frac{\pi}{180}\right) \cos \left(\frac{\pi}{180}(60)\right)\\
&=\frac{\pi}{180} \frac{1}{2}\\
&=\frac{\pi}{360}
\end{align*}
(b) Since
\begin{align*}
\left.\frac{d}{d \theta}(\theta+\tan \theta)\right|_{\theta=45^{\circ}}&=\left.\frac{d}{d \theta}\left(\theta+\tan \left(\frac{\pi}{180} \theta\right)\right)\right|_{\theta=45^{\circ}}\\
&=1+\frac{\pi}{180} \sec ^{2}\left(\frac{\pi}{4}\right)\\
&=1+\frac{\pi}{90}
\end{align*}
