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प्रश्न
If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to
पर्याय
\[- 2x, \frac{1}{2x}\]
\[- \frac{1}{2x}, 2x\]
2x
\[2x, \frac{1}{2x}\]
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उत्तर
\[- 2x, \frac{1}{2x}\]
We have,
\[\tan x = x - \frac{1}{4x}\]
\[ \Rightarrow se c^2 x = 1 + \tan^2 x\]
\[ \Rightarrow se c^2 x = 1 + \left( x - \frac{1}{4x} \right)^2 \]
\[ \Rightarrow se c^2 x = x^2 + \frac{1}{16 x^2} + \frac{1}{2}\]
\[ \Rightarrow se c^2 x = \left( x + \frac{1}{4x} \right)^2 \]
\[ \therefore secx = \pm \left( x + \frac{1}{4x} \right)\]
\[ \Rightarrow secx - \tan x = \left( x + \frac{1}{4x} \right) - \left( x - \frac{1}{4x} \right) or - \left( x + \frac{1}{4x} \right) - \left( x - \frac{1}{4x} \right)\]
\[ = \frac{1}{2x}\text{ or }- 2x\]
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