Recall that #\sec(x)=1/{\cos(x)}#. You can use the formula which states that
#d/{dx} 1/{f(x)} = -\frac{f'}{f^2}#.
This formula can easily be obtained by using the usual formula
#d/{dx} {a(x)}/{b(x)}= \frac{a'\ b - a\ b'}{b^2}#, where #a(x)\equiv 1#, and #b(x)=f(x)#.
Since #d/{dx} \cos(x)=-\sin(x)#, we have that
#d/{dx} 1/{\cos(x)} = -\frac{-\sin(x)}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)}#
For the second derivative of #\sec(x)#, let's derive one more time the first derivative: again, by the rule for the derivation of rational functions, we have
#d/{dx} -\frac{\sin(x)}{\cos^2(x)} = \frac{\sin'(x)\cos^2(x) - \sin(x)(\cos^2(x))'}{\cos^4(x)}#
Since #\sin'(x)=\cos(x)# and #(\cos^2(x))'=-2\cos(x)\sin(x)#, we have
#\frac{\cos^3(x)+2\sin^2(x)\cos(x)}{\cos^4(x)}#
Simplifying #\cos(x)#, we get
#\frac{\cos^2(x)+2\sin^2(x)}{\cos^3(x)}#
Writing #\cos^2(x)# as #1-\sin^2(x)#, we have
#\frac{1-\sin^2(x)+2\sin^2(x)}{\cos^3(x)}=\frac{1+\sin^2(x)}{\cos^3(x)}#
