Question

If $\tan x=x-\frac{1}{4x}$, then $\sec x-\tan x$ is equal to:

• A. $2x$
• B. $-2x$
• C. $4x$
• D. $-4x$

Answer 1

The correct option is B $-2x$

$\tan x=x-\frac{1}{4x}$

We know that,

$se{c}^{2}x=1+ta{n}^{2}x$

$secx=\sqrt{1+ta{n}^{2}x}$

So, $se{c}^{2}x=1+(x-\frac{1}{4x}{)}^2$

$=1+x^2-2.\frac{1}{4x}.x+\frac{1}{16x^2}$

$=x^2+2.\frac{1}{4x}.x+\frac{1}{16x^2}$

$=(x+\frac{1}{4x}{)}^2$

$secx=\pm (x+\frac{1}{4x})$

So either,

$\sec x-\tan x=x+\frac{1}{4x}-(x-\frac{1}{4x})$

$=\frac{1}{2x}$

or, $\sec x-\tan x$

$=-x-\frac{1}{4x}-(x-\frac{1}{4x})$

$=-2x$