If \( \tan x = x – \frac{1}{4x} \), then \( \sec x – \tan x \) is equal to

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

Solution:

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

Hence, correct option is (d).