Question

If sec $\alpha =\frac{2}{\sqrt{3}}$, then find the value of $\left|\frac{1-\text{cosec}\alpha }{1+\text{coesc}\alpha }\right|,$ where $\alpha$ is in IV quadrant.

Answer 1

Given:

$\sec \alpha =\frac{2}{\sqrt{3}}$
$\Rightarrow \sec \alpha =\sec {330}^{\circ }$ (As $\alpha$ is in IV quadrant)
$\alpha ={330}^{\circ }$

$\sin \alpha =\frac{-1}{2}$

$cosec\alpha =-2$
Now,

$\left|\frac{1-\text{cosec}\alpha }{1+\text{coesc}\alpha }\right|$ $=\left|\frac{1-\text{cosec 330}}{1+\text{coesc}330}\right|$

$=\left|\frac{1-(-2)}{1+(-2)}\right|$$=\left|\frac{3}{-1}\right|$$=3$