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Question
If sec \[x = x + \frac{1}{4x}\], then sec x + tan x =
Options
- \[x, \frac{1}{x}\]
- \[2x, \frac{1}{2x}\]
- \[- 2x, \frac{1}{2x}\]
- \[- \frac{1}{x}, x\]
Solution
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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