Answer
(a) is incorrect and (b) is correct.
Work Step by Step
Here, we have
$(\dfrac{1}{\cos x}) \int \cos x dx=(\dfrac{1}{\cos x}) [\sin x+c]$
or, $(\dfrac{1}{\cos x}) \int \cos x dx=\dfrac{\sin x}{\cos x}+\dfrac{c}{\cos x}$
or, $(\dfrac{1}{\cos x}) \int \cos x dx=\tan x+\dfrac{c}{\cos x}$
Thus, it can be seen that solution (a) is incorrect because here the arbitrary constant depends on the values of $c$ but solution (b) is correct because it is the same as the solution.
Hence, (a) is incorrect and (b) is correct.