Question

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

 

Options

• \[x, \frac{1}{x}\]

   

• \[2x, \frac{1}{2x}\]

   

• \[- 2x, \frac{1}{2x}\]

   

• \[- \frac{1}{x}, x\]

   

Answer 1

\[2x, \frac{1}{2x}\]

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