Answer
(a) $0$
(b) $ \ln {(2x+1)}$
(c) $\ln (t-1)$
Work Step by Step
(a) \begin{align*}
\ln \sec \theta+\ln \cos \theta&=\ln \sec \theta \cos \theta\\
&= \ln (1)=0
\end{align*}
(b) \begin{align*}
\ln (8 x+4)-2 \ln 2&=\ln (8 x+4)-\ln 2^2\\
&= \ln \frac{8x+4}{4}\\
&= \ln {(2x+1)}
\end{align*}
(c)
\begin{align*}
3 \ln \sqrt[3]{t^{2}-1}-\ln (t+1)&= \ln {t^{2}-1}-\ln (t+1)\\
&= \ln \frac{ t^2-1}{t+1 }\\
&= \ln \frac{(t-1)(t+1)}{t+1}\\
&= \ln (t-1)
\end{align*}
