MCQ
If $\tan\theta=\sqrt3$ then $\sec\theta=?$
- A$\frac{2}{\sqrt3}$
- B$\frac{\sqrt3}{2}$
- C$\frac{1}{2}$
- ✓$2$
✓
Answer
Correct option: D.
$2$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\tan\theta=\frac{\text{perpendicular}}{\text{Base}}$
$=\frac{\text{BC}}{\text{AB}}=\frac{\sqrt3}{1}$
Let $\text{BC}=\sqrt{3}\text{k}$ and $\text{AB}=\text{k},$
where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(\text{k})^2+(\sqrt3\text{k})^2=\text{k}^2+3\text{k}^2=4\text{k}^2$
$\Rightarrow\text{AC}=\text{2k}$
Now, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$\frac{\text{AC}}{\text{AB}}=\frac{2\text{k}}{\text{k}}=2$
Then, $\tan\theta=\frac{\text{perpendicular}}{\text{Base}}$
$=\frac{\text{BC}}{\text{AB}}=\frac{\sqrt3}{1}$
Let $\text{BC}=\sqrt{3}\text{k}$ and $\text{AB}=\text{k},$
where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(\text{k})^2+(\sqrt3\text{k})^2=\text{k}^2+3\text{k}^2=4\text{k}^2$
$\Rightarrow\text{AC}=\text{2k}$
Now, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$\frac{\text{AC}}{\text{AB}}=\frac{2\text{k}}{\text{k}}=2$
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