Question

If $\text{cosec}\theta =\frac{13}{5}$, then $\tan \theta =......$

• A. $\frac{5}{12}$
• B. $\frac{5}{13}$
• C. $1$
• D. $5$

Answer 1

The correct option is A $\frac{5}{12}$

In $△ABC$, take $m\angle A=\theta ,m\angle B={90}^{\circ }$

$cosec\theta =\frac{AC}{BC}=\frac{13}{5}$

$\therefore$ let $\frac{AC}{13}=\frac{BC}{5}=k$

So, $AC=13k,BC=5k$

By pythagoras, $(13k{)}^2=A{B}^2+(5k{)}^2$

$A{B}^2=169k^2-25k^2=144k^2$,

So, $AB=12k$ ...$(\because k>0)$

So, $AC=13k,BC=5k$ and $AB=12k$

So, by definition of trigonometric ratios,

$\tan \theta =\frac{BC}{AB}=\frac{5k}{12k}=\frac{5}{12}$

So, $\frac{5}{12}$ is correct.