Question

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

• A. $\frac{5}{12}$
• B. $\frac{5}{13}$
• C. $\frac{12}{13}$
• D. None of these

Answer 1

The correct option is C $\frac{12}{13}$

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

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

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

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

By Pythagoras theorem, $13k^2=A{B}^2+5k^2$

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

So, $AB=12k$

By definition of trigonometric ratios,

$\cos \theta =\frac{AB}{AC}=\frac{12k}{13k}=\frac{12}{13}$

So, $\frac{12}{13}$ is correct.