MCQ
If $\tan\theta=\frac{8}{15}$ then $\text{cosec}\theta=?$
- ✓$\frac{17}{8}$
- B$\frac{8}{17}$
- C$\frac{17}{15}$
- D$\frac{15}{17}$
✓
Answer
Correct option: A.
$\frac{17}{8}$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\tan\theta=\frac{\text{perpendicular}}{\text{Base}}$
$=\frac{\text{BC}}{\text{AB}}=\frac{8}{15}$
Let $\text{BC}={8}\text{k}$ and $\text{AB}=15\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(15\text{k})^2+(8\text{k})^2$
$=225\text{k}^2+64\text{k}^2=289\text{k}^2$
$\Rightarrow\text{AC}=\text{17k}$
Now, $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}}$
$\frac{\text{AC}}{\text{BC}}=\frac{17\text{k}}{8\text{k}}=\frac{17}{8}$
Then, $\tan\theta=\frac{\text{perpendicular}}{\text{Base}}$
$=\frac{\text{BC}}{\text{AB}}=\frac{8}{15}$
Let $\text{BC}={8}\text{k}$ and $\text{AB}=15\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(15\text{k})^2+(8\text{k})^2$
$=225\text{k}^2+64\text{k}^2=289\text{k}^2$
$\Rightarrow\text{AC}=\text{17k}$
Now, $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}}$
$\frac{\text{AC}}{\text{BC}}=\frac{17\text{k}}{8\text{k}}=\frac{17}{8}$
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