If \( \sec x = x + \frac{1}{4x} \), then \( \sec x + \tan x = \)

(a) \( x,\ \frac{1}{x} \)
(b) \( 2x,\ \frac{1}{2x} \)
(c) \( -2x,\ \frac{1}{2x} \)
(d) \( -\frac{1}{x},\ x \)

Solution:

\[ \sec x=x+\frac{1}{4x} =\frac{4x^2+1}{4x} \] Using identity, \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] Now, \[ \sec x+\tan x=2x \] Therefore, \[ \sec x-\tan x=\frac{1}{2x} \] Hence, \[ \sec x+\tan x=2x,\qquad \sec x-\tan x=\frac{1}{2x} \]

Hence, correct option is (b).