Answer
$$ H'(t)=\sec t+2\sin t\sec^2 t\tan t$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\sin x)'=\cos x$.
Recall that $(\sec x)'=\sec x\tan x$.
Since $ H(t)=\sin t\sec^2t $, then using the product rule, we have
\begin{align*}
H'(t)&=(\sin t)'\sec^2t+\sin t(\sec^2 t)'\\
&=\cos t\sec^2t+2(\sin t) (\sec t) \sec t \tan t\\
&=\cos t\sec^2t+2\sin t\sec^2 t\tan t\\
&=\sec t+2\sin t\sec^2 t\tan t.
\end{align*}
(Where we used the fact that $\sec t=\dfrac{1}{\cos t}$ in the last line.)
