# If tan x = x

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 $-2 x$