Answer
(a) They are equivalent because we can set $e^{c_1}=c$.
(b) They are equivalent because we can set $c=1+c_1$.
Work Step by Step
(a) We can use the exponentiation rule $e^{x+c_1}=e^{c_1}e^{x}$ and then we can set $c=e^{c_1}$ because if $c_1$ is a constant $e^{c_1}$ is just another constant.
(b) Note that $\sec^2x+c_1=\frac{1}{\cos^2 x}+c_1=\frac{\sin^2 x+\cos^2 x}{\cos^2 x}+c_1 = \frac{\sin^2x}{\cos^2x} +\frac{\cos^2x}{\cos^2x}+c_1 = \tan^2x+1+c_1=\tan^2 x+c$. We can here set $1+c_1=c$, and if $c_1$ is arbitrary constant then $1+c_1$ is also arbitrary constant.
