MCQ
If $\sec\text{x}=\text{x}+\frac{1}{4\text{x}},$ then $\sec\text{x}+\tan\text{x}=$
- A$\text{x},\frac{1}{\text{x}}$
- B$2\text{x},\frac{1}{2\text{x}}$
- C$-2\text{x},\frac{1}{2\text{x}}$
- D$-\frac{1}{\text{x}},\text{x}$
Solution:
We have:
$\sec\text{x} = \text{x} +\frac{1}{4\text{x}}$
$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$
$\Rightarrow1+\tan^2\text{x}$
$=1+\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$
$\therefore\tan\text{x}=\pm\Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$\sec\text{x}-\tan\text{x}=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{or}$
$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big[-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\Big]$
$=\frac{1}{2\text{x}}\text{ or } 2\text{x}$
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