# If Sec $X = X + \Frac{1}{4x}$, Then Sec X + Tan X = - Mathematics

MCQ

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

#### Options

• $x, \frac{1}{x}$

• $2x, \frac{1}{2x}$

• $- 2x, \frac{1}{2x}$

• $- \frac{1}{x}, x$

#### Solution

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

We have,
$secx = x + \frac{1}{4x}$
$\Rightarrow se c^2 x = = x^2 + \frac{1}{16 x^2} + \frac{1}{2}$
$\Rightarrow 1 + \tan^2 x = 1 + x^2 + \frac{1}{16 x^2} - \frac{1}{2}$
$\Rightarrow \tan^2 x = x^2 + \frac{1}{16 x^2} - \frac{1}{2}$
$\Rightarrow \tan^2 x = \left( x - \frac{1}{4x} \right)^2$
$\therefore \tan x = \pm \left( x - \frac{1}{4x} \right)$
$\Rightarrow sec x - \tan x = \left( x + \frac{1}{4x} \right) - \left( x - \frac{1}{4x} \right) or \left( x + \frac{1}{4x} \right) - \left[ - \left( x - \frac{1}{4x} \right) \right]$
$= \frac{1}{2x}\text{ or }2x$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Q 2 | Page 41