Question:
If $\tan x=x-\frac{1}{4 x}$, then $\sec x-\tan x$ is equal to
(a) $-2 x, \frac{1}{2 x}$
(b) $-\frac{1}{2 x}, 2 x$
(c) $2 x$
(d) $2 x, \frac{1}{2 x}$
Solution:
(a) $-2 x, \frac{1}{2 x}$
We have,
$\tan x=x-\frac{1}{4 x}$
$\Rightarrow \sec ^{2} x=1+\tan ^{2} x$
$\Rightarrow \sec ^{2} x=1+\left(x-\frac{1}{4 x}\right)^{2}$
$\Rightarrow \sec ^{2} x=x^{2}+\frac{1}{16 x^{2}}+\frac{1}{2}$
$\Rightarrow \sec ^{2} x=\left(x+\frac{1}{4 x}\right)^{2}$
$\therefore \sec x=\pm\left(x+\frac{1}{4 x}\right)$
$\Rightarrow \sec x-\tan x=\left(x+\frac{1}{4 x}\right)-\left(x-\frac{1}{4 x}\right)$ or $-\left(x+\frac{1}{4 x}\right)-\left(x-\frac{1}{4 x}\right)$
$=\frac{1}{2 x}$ or $-2 x$