MCQ
If $\text{cosec}\theta=\sqrt{10}$ then $\sec\theta=?$
- A$\frac{3}{\sqrt{10}}$
- ✓$\frac{\sqrt{10}}{3}$
- C$\frac{1}{\sqrt{10}}$
- D$\frac{2}{\sqrt{10}}$
✓
Answer
Correct option: B.
$\frac{\sqrt{10}}{3}$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{perpendicular}}$
$=\frac{\text{AC}}{\text{BC}}=\frac{\sqrt{10}}{1}$
Let $\text{AC}=\sqrt{10}\text{k}$ and $\text{BC}=\text{k}$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\big(\sqrt{10}\text{k}\big)^2=\text{AB}^2+(\text{k})^2$
$\Rightarrow\text{AB}^2=10\text{k}^2-\text{k}^2=\text{9k}^2$
$\Rightarrow\text{AB}=\text{3k}$
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}$
$=\frac{\sqrt{10}\text{k}}{\text{3k}}=\frac{\sqrt{10}}3{}$
Then, $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{perpendicular}}$
$=\frac{\text{AC}}{\text{BC}}=\frac{\sqrt{10}}{1}$
Let $\text{AC}=\sqrt{10}\text{k}$ and $\text{BC}=\text{k}$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\big(\sqrt{10}\text{k}\big)^2=\text{AB}^2+(\text{k})^2$
$\Rightarrow\text{AB}^2=10\text{k}^2-\text{k}^2=\text{9k}^2$
$\Rightarrow\text{AB}=\text{3k}$
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}$
$=\frac{\sqrt{10}\text{k}}{\text{3k}}=\frac{\sqrt{10}}3{}$
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