Question

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

Answer 1

Since, $\alpha$ is in the IV th quadrant, so, $\text{cosec}\alpha$ is negative.

We have,

$\sec \alpha =\frac{2}{\sqrt{3}}$
So, $\cos \alpha =\frac{1}{\sec \alpha }$ $=\frac{\sqrt{3}}{2}$
Now,

${\sin }^2\alpha +{\cos }^2\alpha =1$
${\sin }^2\alpha =1-{\cos }^2\alpha$ $=1-\frac{3}{4}=\frac{1}{4}$
$\sin \alpha =\sqrt{\frac{1}{4}}=\pm \frac{1}{2}$
$\sin \alpha =-\frac{1}{2}$ ($\because \alpha$ is in $IV$ quadrant)
$\therefore \text{cosec}\alpha =\frac{1}{\sin \alpha }=-2$
Now,

$\frac{1-\text{cosec}\alpha }{1+\text{cosec}\alpha }=\frac{1-(-2)}{1+(-2)}$ $=\frac{3}{-1}$ $=-3$