Question

If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to

Options

• \[- 2x, \frac{1}{2x}\]

• \[- \frac{1}{2x}, 2x\]

• 2x

• \[2x, \frac{1}{2x}\]

Answer 1

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