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