Answer
$$f'(t)=-\sec^2(\sec(\cos t))\sec(\cos t)\tan(\cos t)\sin t$$
Work Step by Step
$$f'(t)=(\tan(\sec(\cos t)))'=\frac{1}{\cos^2(\sec(\cos t))}(\sec(\cos t))'=
\sec^2(\sec(\cos t))\left(\frac{1}{\cos(\cos t)}\right)'=
\sec^2(\sec(\cos t))(-\frac{1}{\cos^2(\cos t)})(\cos(\cos t))'=
-\sec^2(\sec(\cos t))\frac{1}{\cos^2(\cos t))}(-\sin(\cos t))(\cos t)'=
\sec^2(\sec(\cos t))\sec(\cos t)\tan(\cos t)(-\sin t)=
-\sec^2(\sec(\cos t))\sec(\cos t)\tan(\cos t)\sin t$$
