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Trigonometric Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\tan\text{x}=\text{x}+\frac{1}{4\text{x}},$ then $\sec\text{x}+\tan\text{x}=$
  • ✓
    $-2\text{x},\frac{1}{2\text{x}}$
  • B
    $-\frac{1}{2\text{x}},2\text{x}$
  • C
    $2\text{x}$
  • D
    $2\text{x},\frac{1}{\text{x}2}$
✓

Answer

Correct option: A.
$-2\text{x},\frac{1}{2\text{x}}$
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} $
$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)- \Big(\text{x}-\frac{1}{4\text{x}}\Big)$ or $ -\Big(\text{x}+\frac{1}{4\text{x}}\Big)$ or $ - \Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$=\frac{1}{2\text{x}}$ or $-2\text{x}$

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